Rs Aggarwal 2020 2021 for Class 9 Maths Chapter 2 - Polynomials

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Rs Aggarwal 2020 2021 Solutions for Class 9 Maths Chapter 2 Polynomials are provided here with simple step-by-step explanations. These solutions for Polynomials are extremely popular among Class 9 students for Maths Polynomials Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rs Aggarwal 2020 2021 Book of Class 9 Maths Chapter 2 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rs Aggarwal 2020 2021 Solutions. All Rs Aggarwal 2020 2021 Solutions for class Class 9 Maths are prepared by experts and are 100% accurate.

Page No 74:

Answer:

(i) $x^{5} - 2 x^{3} + x + \sqrt{3}$ is an expression having only non-negative integral powers of x. So, it is a polynomial. Also, the highest power of x is 5, so, it is a polynomial of degree 5.

(ii) $y^{3} + \sqrt{3} y$ is an expression having only non-negative integral powers of y. So, it is a polynomial. Also, the highest power of y is 3, so, it is a polynomial of degree 3.

(iii) $t^{2} - \frac{2}{5} t + \sqrt{5}$ is an expression having only non-negative integral powers of t. So, it is a polynomial. Also, the highest power of t is 2, so, it is a polynomial of degree 2.

(iv) $x^{100} - 1$ is an expression having only non-negative integral power of x. So, it is a polynomial. Also, the highest power of x is 100, so, it is a polynomial of degree 100.

(v) $\frac{1}{\sqrt{2}} x^{2} - \sqrt{2} x + 2$ is an expression having only non-negative integral powers of x. So, it is a polynomial. Also, the highest power of x is 2, so, it is a polynomial of degree 2.

(vi) $x^{- 2} + 2 x^{- 1} + 3$ is an expression having negative integral powers of x. So, it is not a polynomial.

(vii) Clearly, 1 is a constant polynomial of degree 0.

(viii) Clearly, $- \frac{3}{5}$ is a constant polynomial of degree 0.

(ix) $\frac{x^{2}}{2} - \frac{2}{x^{2}} = \frac{x^{2}}{2} - 2 x^{- 2}$
This is an expression having negative integral power of x i.e. −2. So, it is not a polynomial.

(x) $\sqrt[3]{2} x^{2} - 8$ is an expression having only non-negative integral power of x. So, it is a polynomial. Also, the highest power of x is 2, so, it is a polynomial of degree 2.

(xi) $\frac{1}{2 x^{2}} = \frac{1}{2} x^{- 2}$ is an expression having negative integral power of x. So, it is not a polynomial.

(xii) $\frac{1}{\sqrt{5}} x^{\frac{1}{2}} + 1$
In this expression, the power of x is $\frac{1}{2}$ which is a fraction. Since it is an expression having fractional power of x, so, it is not a polynomial.

(xiii) $\frac{3}{5} x^{2} - \frac{7}{3} x + 9$ is an expression having only non-negative integral powers of x. So, it is a polynomial. Also, the highest power of x is 2, so, it is a polynomial of degree 2.

(xiv) $x^{4} - x^{\frac{3}{2}} + x - 3$
In this expression, one of the powers of x is $\frac{3}{2}$ which is a fraction. Since it is an expression having fractional power of x, so, it is not a polynomial.

(xv) $2 x^{3} + 3 x^{2} + \sqrt{x} - 1 = 2 x^{3} + 3 x^{2} + x^{\frac{1}{2}} - 1$
In this expression, one of the powers of x is $\frac{1}{2}$ which is a fraction. Since it is an expression having fractional power of x, so, it is not a polynomial.

Page No 75:

Answer:

(i) –7 + x is a polynomial with degree 1. So, it is a linear polynomial.

(ii) 6y is a polynomial with degree 1. So, it is a linear polynomial.

(iii) –z³ is a polynomial with degree 3. So, it is a cubic polynomial.

(iv) 1 – y – y³ is a polynomial with degree 3. So, it is a cubic polynomial.

(v) x – x³ + x⁴ is a polynomial with degree 4. So, it is a quartic polynomial.

(vi) 1 + x + x² is a polynomial with degree 2. So, it is a quadratic polynomial.

(vii) – 6x² is a polynomial with degree 2. So, it is a quadratic polynomial.

(viii) –13 is a polynomial with degree 0. So, it is a constant polynomial.

(ix) – p is a polynomial with degree 1. So, it is a linear polynomial.

Page No 75:

Answer:

(i) The coefficient of x³ in $x + 3 x^{2} - 5 x^{3} + x^{4}$ is −5.

(ii) The coefficient of x in $\sqrt{3} - 2 \sqrt{2} x + 6 x^{2}$ is $- 2 \sqrt{2}$ .

(iii) 2x – 3 + x³= – 3 + 2x + 0x²+ x³
The coefficient of x² in 2x – 3 + x³ is 0.

(iv) The coefficient of x in $\frac{3}{8} x^{2} - \frac{2}{7} x + \frac{1}{6}$ is $- \frac{2}{7}$ .

(v) The constant term in $\frac{π}{2} x^{2} + 7 x - \frac{2}{5} π$ is $- \frac{2}{5} π$ .

Page No 75:

Answer:

(i) $\frac{4 x - 5 x^{2} + 6 x^{3}}{2 x} = \frac{4 x}{2 x} - \frac{5 x^{2}}{2 x} + \frac{6 x^{3}}{2 x} = 2 - \frac{5}{2} x + 3 x^{2}$
Here, the highest power of x is 2. So, the degree of the polynomial is 2.

(ii) y²(y – y³) = y³ – y⁵
Here, the highest power of y is 5. So, the degree of the polynomial is 5.

(iii) (3x – 2)(2x³ + 3x²) = 6x⁴ + 9x³ – 4x³ – 6x² = 6x⁴ + 5x³ – 6x²
Here, the highest power of x is 4. So, the degree of the polynomial is 4.

(iv) $- \frac{1}{2} x + 3$
Here, the highest power of x is 1. So, the degree of the polynomial is 1.

(v) – 8
–8 is a constant polynomial. So, the degree of the polynomial is 0.

(vi) x^–2(x⁴+ x²) = x²+ x⁰= x²+ 1
Here, the highest power of x is 2. So, the degree of the polynomial is 2.

Page No 75:

Answer:

(i) A polynomial having one term is called a monomial. Since the degree of required monomial is 5, so the highest power of x in the monomial should be 5.
An example of a monomial of degree 5 is 2x⁵.

(ii) A polynomial having two terms is called a binomial. Since the degree of required binomial is 8, so the highest power of x in the binomial should be 8.
An example of a binomial of degree 8 is 2x⁸ − 3x.

(iii) A polynomial having three terms is called a trinomial. Since the degree of required trinomial is 4, so the highest power of x in the trinomial should be 4.
An example of a trinomial of degree 4 is 2x⁴ − 3x + 5.

(iv) A polynomial having one term is called a monomial. Since the degree of required monomial is 0, so the highest power of x in the monomial should be 0.
An example of a monomial of degree 0 is 5.

Page No 75:

Answer:

A polynomial written either in ascending or descending powers of a variable is called the standard form of a polynomial.

(i) $8 + x - 2 x^{2} + 5 x^{3}$ is a polynomial in standard form as the powers of x are in ascending order.
(ii) $\frac{2}{3} - 3 y + 4 y^{2} + 2 y^{3}$ is a polynomial in standard form as the powers of y are in ascending order.
(iii) $2 x - 3 x^{2} + 6 x^{3} - x^{5}$ is a polynomial in standard form as the powers of x are in ascending order.
(iv) $2 + t - t^{2} - 3 t^{3} + t^{4}$ is a polynomial in standard form as the powers of t are in ascending order.

Page No 78:

Answer:

$(i) p (x) = 5 - 4 x + 2 x^{2} \Rightarrow p (0) = (5 - 4 \times 0 + 2 \times 0^{2})$
$= (5 - 0 + 0) = 5$

$(ii) p (x) = 5 - 4 x + 2 x^{2} \Rightarrow p (3) = (5 - 4 \times 3 + 2 \times 3^{2})$
$= (5 - 12 + 18) = 11$

$(iii) p (x) = 5 - 4 x + 2 x^{2} \Rightarrow p (- 2) = [5 - 4 \times (- 2) + 2 \times {(- 2)}^{2}]$
$= (5 + 8 + 8) = 21$

Page No 78:

Answer:

$(i) p (y) = 4 + 3 y - y^{2} + 5 y^{3} \Rightarrow p (0) = (4 + 3 \times 0 - 0^{2} + 5 \times 0^{3})$
$= (4 + 0 - 0 + 0) = 4$

$(ii) p (y) = 4 + 3 y - y^{2} + 5 y^{3} \Rightarrow p (2) = (4 + 3 \times 2 - 2^{2} + 5 \times 2^{3})$
$= (4 + 6 - 4 + 40) = 46$

$(iii) p (y) = 4 + 3 y - y^{2} + 5 y^{3} \Rightarrow p (- 1) = [4 + 3 \times (- 1) - {(- 1)}^{2} + 5 \times {(- 1)}^{3}]$
$= (4 - 3 - 1 - 5) = - 5$

Page No 78:

Answer:

$(i) f (t) = 4 t^{2} - 3 t + 6 \Rightarrow f (0) = (4 \times 0^{2} - 3 \times 0 + 6)$
$= (0 - 0 + 6) = 6$

$(ii) f (t) = 4 t^{2} - 3 t + 6 \Rightarrow f (4) = (4 \times 4^{2} - 3 \times 4 + 6)$
$= (64 - 12 + 6) = 58$

$(iii) f (t) = 4 t^{2} - 3 t + 6 \Rightarrow f (- 5) = [4 \times {(- 5)}^{2} - 3 \times (- 5) + 6]$
$= (100 + 15 + 6) = 121$

Page No 78:

Answer:

$p (x) = x^{3} - 3 x^{2} + 2 x$ .....(1)

Putting x = 0 in (1), we get

$p (0) = 0^{3} - 3 \times 0^{2} + 2 \times 0 = 0$

Thus, x = 0 is a zero of p(x).

Putting x = 1 in (1), we get

$p (1) = 1^{3} - 3 \times 1^{2} + 2 \times 1 = 1 - 3 + 2 = 0$

Thus, x = 1 is a zero of p(x).

Putting x = 2 in (1), we get

$p (2) = 2^{3} - 3 \times 2^{2} + 2 \times 2 = 8 - 3 \times 4 + 4 = 8 - 12 + 4 = 0$

Thus, x = 2 is a zero of p(x).

Page No 78:

Answer:

p(x) = x³ + x² – 9x – 9 .....(1)

Putting x = 0 in (1), we get

p(0) = 0³ + 0² – 9 × 0 – 9 = 0 + 0 – 0 – 9 = –9 ≠ 0

Thus, x = 0 is not a zero of p(x).

Putting x = 3 in (1), we get

p(3) = 3³ + 3² – 9 × 3 – 9 = 27 + 9 – 27 – 9 = 0

Thus, x = 3 is a zero of p(x).

Putting x = –3 in (1), we get

p(–3) = (–3)³ + (–3)² – 9 × (–3) – 9 = –27 + 9 + 27 – 9 = 0

Thus, x = –3 is a zero of p(x).

Putting x = –1 in (1), we get

p(–1) = (–1)³ + (–1)² – 9 × (–1) – 9 = –1 + 1 + 9 – 9 = 0

Thus, x = –1 is a zero of p(x).

Page No 78:

Answer:

$(i) p (x) = x - 4 \Rightarrow p (4) = 4 - 4$
= 0
Hence, 4 is the zero of the given polynomial.

$(ii) p (x) = (- 3) + 3 \Rightarrow p (3) = 0$
Hence, 3 is the zero of the given polynomial.

$(iii) p (x) = 2 - 5 x \Rightarrow p (\frac{2}{5}) = 2 - 5 \times (\frac{2}{5})$
$= 2 - 2 = 0$

Hence, $\frac{2}{5}$ is the zero of the given polynomial.

$(iv) p (y) = 2 y + 1 \Rightarrow p (- \frac{1}{2}) = 2 \times (- \frac{1}{2}) + 1 = - 1 + 1 = 0$

Hence, $- \frac{1}{2}$ is the zero of the given polynomial.

Page No 79:

Answer:

$(i) p (x) = x^{2} - 3 x + 2 = (x - 1) (x - 2) \Rightarrow p (1) = (1 - 1) \times (1 - 2)$
$= 0 \times (- 1) = 0$
Also,
$p (2) = (2 - 1) (2 - 2)$
$= (- 1) \times 0 = 0$

Hence, 1 and 2 are the zeroes of the given polynomial.

$(ii) p (x) = x^{2} + x - 6 \Rightarrow p (2) = 2^{2} + 2 - 6$
$= 4 - 4 = 0$
Also,
$p (- 3) = {(- 3)}^{2} + (- 3) - 6 = 9 - 9 = 0$

Hence, 2 and $- 3$ are the zeroes of the given polynomial.

$(iii) p (x) = x^{2} - 3 x \Rightarrow p (0) = 0^{2} - 3 \times 0$

Also,
$p (3) = 3^{2} - 3 \times 3 = 9 - 9 = 0$

Hence, 0 and 3 are the zeroes of the given polynomial.

Page No 79:

Answer:

$(i) p (x) = 0 \Rightarrow x - 5 = 0 \Rightarrow x = 5 Hence, 5 is the zero of the polynomial p (x) . (ii) q (x) = 0 \Rightarrow x + 4 = 0 \Rightarrow x = - 4 Hence, - 4 is the zero of the polynomial q (x) . (iii) r (x) = 0 \Rightarrow 2 x + 5 = 0 \Rightarrow t = \frac{- 5}{2} Hence, \frac{- 5}{2} is the zero of the polynomial p (t) .$

$(iv) f (x) = 0 \Rightarrow 3 x + 1 = 0 \Rightarrow x = - \frac{1}{3} Hence, - \frac{1}{3} is the zero of the polynomial f (x) . (v) g (x) = 0 \Rightarrow 5 - 4 x = 0 \Rightarrow x = \frac{5}{4} Hence, \frac{5}{4} is the zero of the polynomial g (x) . (vi) h (x) = 0 \Rightarrow 6 x - 2 = 0 \Rightarrow x = \frac{2}{6} = \frac{1}{3} Hence, \frac{1}{3} is the zero of the polynomial h (x) .$

$(vii) p (x) = 0 \Rightarrow a x = 0 \Rightarrow x = 0 Hence, 0 is the zero of the polynomial p (x) . (viii) q (x) = 0 \Rightarrow 4 x = 0 \Rightarrow x = 0 Hence, 0 is the zero of the polynomial q (x) .$

Page No 79:

Answer:

It is given that 2 and 0 are the zeroes of the polynomial $f (x) = 2 x^{3} - 5 x^{2} + a x + b$ .
∴ f(2) = 0
$\Rightarrow 2 \times 2^{3} - 5 \times 2^{2} + a \times 2 + b = 0 \Rightarrow 16 - 20 + 2 a + b = 0 \Rightarrow - 4 + 2 a + b = 0 \Rightarrow 2 a + b = 4 . . . . . (1)$
Also,
f(0) = 0
$\Rightarrow 2 \times 0^{3} - 5 \times 0^{2} + a \times 0 + b = 0 \Rightarrow 0 - 0 + 0 + b = 0 \Rightarrow b = 0$
Putting b = 0 in (1), we get
$2 a + 0 = 4 \Rightarrow 2 a = 4 \Rightarrow a = 2$
Thus, the values of a and b are 2 and 0, respectively.

Page No 84:

Answer:

Let f(x) = x⁴ + 1 and g(x) = x – 1.

Quotient = x³ + x² + x + 1

Remainder = 2

Verification:

Putting x = 1 in f(x), we get

f(1) = 1⁴ + 1 = 1 + 1 = 2 = Remainder, when f(x) = x⁴ + 1 is divided by g(x) = x – 1

Page No 84:

Answer:

$p (x) = 2 x^{4} - 6 x^{3} + 2 x^{2} - x + 2$ and $g (x) = x + 2$

Quotient = $2 x^{3} - 10 x^{2} + 22 x - 45$

Remainder = 92

Verification:

Divisor × Quotient + Remainder
$= (x + 2) \times (2 x^{3} - 10 x^{2} + 22 x - 45) + 92 = x (2 x^{3} - 10 x^{2} + 22 x - 45) + 2 (2 x^{3} - 10 x^{2} + 22 x - 45) + 92 = 2 x^{4} - 10 x^{3} + 22 x^{2} - 45 x + 4 x^{3} - 20 x^{2} + 44 x - 90 + 92 = 2 x^{4} - 6 x^{3} + 2 x^{2} - x + 2$
= Dividend

Hence verified.

Page No 84:

Answer:

$p (x) = x^{3} - 6 x^{2} + 9 x + 3$

$g (x) = x - 1$

By remainder theorem, when p(x) is divided by (x − 1), then the remainder = p(1).

Putting x = 1 in p(x), we get

$p (1) = 1^{3} - 6 \times 1^{2} + 9 \times 1 + 3 = 1 - 6 + 9 + 3 = 7$

∴ Remainder = 7

Thus, the remainder when p(x) is divided by g(x) is 7.

Page No 84:

Answer:

$p (x) = 2 x^{3} - 7 x^{2} + 9 x - 13$

$g (x) = x - 3$

By remainder theorem, when p(x) is divided by (x − 3), then the remainder = p(3).

Putting x = 3 in p(x), we get

$p (3) = 2 \times 3^{3} - 7 \times 3^{2} + 9 \times 3 - 13 = 54 - 63 + 27 - 13 = 5$

∴ Remainder = 5

Thus, the remainder when p(x) is divided by g(x) is 5.

Page No 84:

Answer:

$p (x) = 3 x^{4} - 6 x^{2} - 8 x - 2$

$g (x) = x - 2$

By remainder theorem, when p(x) is divided by (x − 2), then the remainder = p(2).

Putting x = 2 in p(x), we get

$p (2) = 3 \times 2^{4} - 6 \times 2^{2} - 8 \times 2 - 2 = 48 - 24 - 16 - 2 = 6$

∴ Remainder = 6

Thus, the remainder when p(x) is divided by g(x) is 6.

Page No 84:

Answer:

$p (x) = 2 x^{3} - 9 x^{2} + x + 15$

$g (x) = 2 x - 3 = 2 (x - \frac{3}{2})$

By remainder theorem, when p(x) is divided by (2x − 3), then the remainder = $p (\frac{3}{2})$ .

Putting $x = \frac{3}{2}$ in p(x), we get

$p (\frac{3}{2}) = 2 \times {(\frac{3}{2})}^{3} - 9 \times {(\frac{3}{2})}^{2} + \frac{3}{2} + 15 = \frac{27}{4} - \frac{81}{4} + \frac{3}{2} + 15$ $= \frac{27 - 81 + 6 + 60}{4} = \frac{12}{4} = 3$

∴ Remainder = 3

Thus, the remainder when p(x) is divided by g(x) is 3.

Page No 84:

Answer:

$p (x) = x^{3} - 2 x^{2} - 8 x - 1$

$g (x) = x + 1$

By remainder theorem, when p(x) is divided by (x + 1), then the remainder = p(−1).

Putting x = −1 in p(x), we get

$p (- 1) = {(- 1)}^{3} - 2 \times {(- 1)}^{2} - 8 \times (- 1) - 1 = - 1 - 2 + 8 - 1 = 4$

∴ Remainder = 4

Thus, the remainder when p(x) is divided by g(x) is 4.

Page No 84:

Answer:

$p (x) = 2 x^{3} + x^{2} - 15 x - 12$

$g (x) = x + 2$

By remainder theorem, when p(x) is divided by (x + 2), then the remainder = p(−2).

Putting x = −2 in p(x), we get

$p (- 2) = 2 \times {(- 2)}^{3} + {(- 2)}^{2} - 15 \times (- 2) - 12 = - 16 + 4 + 30 - 12 = 6$

∴ Remainder = 6

Thus, the remainder when p(x) is divided by g(x) is 6.

Page No 84:

Answer:

$p (x) = 6 x^{3} + 13 x^{2} + 3$

$g (x) = 3 x + 2 = 3 (x + \frac{2}{3}) = 3 [x - (- \frac{2}{3})]$

By remainder theorem, when p(x) is divided by (3x + 2), then the remainder = $p (- \frac{2}{3})$ .

Putting $x = - \frac{2}{3}$ in p(x), we get

$p (- \frac{2}{3}) = 6 \times {(- \frac{2}{3})}^{3} + 13 \times {(- \frac{2}{3})}^{2} + 3 = - \frac{16}{9} + \frac{52}{9} + 3$ $= \frac{- 16 + 52 + 27}{9} = \frac{63}{9} = 7$

∴ Remainder = 7

Thus, the remainder when p(x) is divided by g(x) is 7.

Page No 84:

Answer:

$p (x) = x^{3} - 6 x^{2} + 2 x - 4$

$g (x) = 1 - \frac{3}{2} x = - \frac{3}{2} (x - \frac{2}{3})$

By remainder theorem, when p(x) is divided by $(1 - \frac{3}{2} x)$ , then the remainder = $p (\frac{2}{3})$ .

Putting $x = \frac{2}{3}$ in p(x), we get

$p (\frac{2}{3}) = {(\frac{2}{3})}^{3} - 6 \times {(\frac{2}{3})}^{2} + 2 \times (\frac{2}{3}) - 4 = \frac{8}{27} - \frac{8}{3} + \frac{4}{3} - 4$ $= \frac{8 - 72 + 36 - 108}{27} = - \frac{136}{27}$

∴ Remainder = $- \frac{136}{27}$

Thus, the remainder when p(x) is divided by g(x) is $- \frac{136}{27}$ .

Page No 84:

Answer:

$p (x) = 2 x^{3} + 3 x^{2} - 11 x - 3$

$g (x) = (x + \frac{1}{2}) = [x - (- \frac{1}{2})]$

By remainder theorem, when p(x) is divided by $(x + \frac{1}{2})$ , then the remainder = $p (- \frac{1}{2})$ .

Putting $x = - \frac{1}{2}$ in p(x), we get

$p (- \frac{1}{2}) = 2 \times {(- \frac{1}{2})}^{3} + 3 \times {(- \frac{1}{2})}^{2} - 11 \times (- \frac{1}{2}) - 3 = - \frac{1}{4} + \frac{3}{4} + \frac{11}{2} - 3$ $= \frac{- 1 + 3 + 22 - 12}{4} = \frac{12}{4} = 3$

∴ Remainder = 3

Thus, the remainder when p(x) is divided by g(x) is 3.

Page No 84:

Answer:

$p (x) = x^{3} - a x^{2} + 6 x - a$

$g (x) = x - a$

By remainder theorem, when p(x) is divided by (x − a), then the remainder = p(a).

Putting x = a in p(x), we get

$p (a) = a^{3} - a \times a^{2} + 6 \times a - a = a^{3} - a^{3} + 6 a - a = 5 a$

∴ Remainder = 5a

Thus, the remainder when p(x) is divided by g(x) is 5a.

Page No 84:

Answer:

Let $f (x) = 2 x^{3} + x^{2} - a x + 2$ and $g (x) = 2 x^{3} - 3 x^{2} - 3 x + a$ .

By remainder theorem, when f(x) is divided by (x – 2), then the remainder = f(2).

Putting x = 2 in f(x), we get

$f (2) = 2 \times 2^{3} + 2^{2} - a \times 2 + 2 = 16 + 4 - 2 a + 2 = - 2 a + 22$

By remainder theorem, when g(x) is divided by (x – 2), then the remainder = g(2).

Putting x = 2 in g(x), we get

$g (2) = 2 \times 2^{3} - 3 \times 2^{2} - 3 \times 2 + a = 16 - 12 - 6 + a = - 2 + a$

It is given that,

$f (2) = g (2) \Rightarrow - 2 a + 22 = - 2 + a \Rightarrow - 3 a = - 24 \Rightarrow a = 8$
Thus, the value of a is 8.

Page No 84:

Answer:

$Let : p (x) = x^{4} - 2 x^{3} + 3 x^{2} - a x + b$

$Now, When p (x) is divided by (x - 1), the remainder is p (1) . When p (x) is divided by (x + 1), the remainder is p (- 1) .$
Thus, we have:
$p (1) = (1^{4} - 2 \times 1^{3} + 3 \times 1^{2} - a \times 1 + b) = (1 - 2 + 3 - a + b) = 2 - a + b$
And,
$p (- 1) = [{(- 1)}^{4} - 2 \times {(- 1)}^{3} + 3 \times {(- 1)}^{2} - a \times (- 1) + b] = (1 + 2 + 3 + a + b) = 6 + a + b$

Now,

$2 - a + b = 5 . . . (1) 6 + a + b = 19 . . . (2)$

$Adding (1) and (2), we get : 8 + 2 b = 24$
$\Rightarrow 2 b = 16 \Rightarrow b = 8$
By putting the value of b, we get the value of a, i.e., 5.
∴ a = 5 and b = 8
Now,
$f (x) = x^{4} - 2 x^{3} + 3 x^{2} - 5 x + 8$
Also,
$When p (x) is divided by (x - 2), the remainder is p (2) .$
Thus, we have:
$p (2) = (2^{4} - 2 \times 2^{3} + 3 \times 2^{2} - 5 \times 2 + 8) [a = 5 and b = 8] = (16 - 16 + 12 - 10 + 8) = 10$

Page No 85:

Answer:

$p (x) = x^{3} - 5 x^{2} + 4 x - 3$

$g (x) = x - 2$

Putting x = 2 in p(x), we get

$p (2) = 2^{3} - 5 \times 2^{2} + 4 \times 2 - 3 = 8 - 20 + 8 - 3 = - 7 \neq 0$

Therefore, by factor theorem, (x − 2) is not a factor of p(x).

Hence, p(x) is not a multiple of g(x).

Page No 85:

Answer:

$p (x) = 2 x^{3} - 11 x^{2} - 4 x + 5$

$g (x) = 2 x + 1 = 2 (x + \frac{1}{2}) = 2 [x - (- \frac{1}{2})]$

Putting $x = - \frac{1}{2}$ in p(x), we get

$p (- \frac{1}{2}) = 2 \times {(- \frac{1}{2})}^{3} - 11 \times {(- \frac{1}{2})}^{2} - 4 \times (- \frac{1}{2}) + 5 = - \frac{1}{4} - \frac{11}{4} + 2 + 5$ $= - \frac{12}{4} + 7 = - 3 + 7 = 4 \neq 0$

Therefore, by factor theorem, (2x + 1) is not a factor of p(x).

Hence, g(x) is not a factor of p(x).

Page No 90:

Answer:

Let:
p(x) = x³ – 8
Now,
$g (x) = 0 \Rightarrow x - 2 = 0 \Rightarrow x = 2$
By the factor theorem, (x – 2) is a factor of the given polynomial if p(2) = 0.
Thus, we have:
$p (2) = (2^{3} - 8) = 0$
Hence, (x $-$ 2) is a factor of the given polynomial.

Page No 90:

Answer:

Let:
p(x) = 2x³ + 7x² – 24x – 45
Now,
$x - 3 = 0 \Rightarrow x = 3$
By the factor theorem, (x $-$ 3) is a factor of the given polynomial if p(3) = 0.
Thus, we have:
$p (3) = (2 \times 3^{3} - 7 \times 3^{2} - 24 \times 3 - 45) = (54 + 63 - 72 - 45) = 0$
Hence, (x $-$ 3) is a factor of the given polynomial.

Page No 90:

Answer:

Let:
p(x) = 2x⁴ + 9x³ + 6x² – 11x – 6
Here,
$x - 1 = 0 \Rightarrow x = 1$
By the factor theorem, (x $-$ 1) is a factor of the given polynomial if p(1) = 0.
Thus, we have:
$p (1) = (2 \times 1^{4} + 9 \times 1^{3} + 6 \times 1^{2} - 11 \times 1 - 6) = (2 + 9 + 6 - 11 - 6) = 0$
Hence, (x $-$ 1) is a factor of the given polynomial.

Page No 91:

Answer:

Let:
p(x) = x⁴ – x² – 12
Here,
$x + 2 = 0 \Rightarrow x = - 2$
By the factor theorem, (x + 2) is a factor of the given polynomial if p ( $-$ 2) = 0.
Thus, we have:
$p (- 2) = [{(- 2)}^{4} - {(- 2)}^{2} - 12] = (16 - 4 - 12) = 0$
Hence, (x + 2) is a factor of the given polynomial.

Page No 91:

Answer:

$p (x) = 69 + 11 x - x^{2} + x^{3}$

$g (x) = x + 3$

Putting x = −3 in p(x), we get

$p (- 3) = 69 + 11 \times (- 3) - {(- 3)}^{2} + {(- 3)}^{3} = 69 - 33 - 9 - 27 = 0$

Therefore, by factor theorem, (x + 3) is a factor of p(x).

Hence, g(x) is a factor of p(x).

Page No 91:

Answer:

Let:
$p (x) = 2 x^{3} + 9 x^{2} - 11 x - 30$
Here,
$x + 5 = 0 \Rightarrow x = - 5$
By the factor theorem, (x + 5) is a factor of the given polynomial if p ( $-$ 5) = 0.
Thus, we have:
$p (- 5) = [2 \times {(- 5)}^{3} + 9 \times {(- 5)}^{2} - 11 \times (- 5) - 30] = (- 250 + 225 + 55 - 30) = 0$
Hence, (x + 5) is a factor of the given polynomial.

Page No 91:

Answer:

Let:
$p (x) = 2 x^{4} + x^{3} - 8 x^{2} - x + 6$
Here,
$2 x - 3 = 0 \Rightarrow x = \frac{3}{2}$
By the factor theorem, (2x $-$ 3) is a factor of the given polynomial if $p (\frac{3}{2}) = 0$ .
Thus, we have:

$p (\frac{3}{2}) = [2 \times {(\frac{3}{2})}^{4} + {(\frac{3}{2})}^{3} - 8 \times {(\frac{3}{2})}^{2} - (\frac{3}{2}) + 6] = (\frac{81}{8} + \frac{27}{8} - 18 - \frac{3}{2} + 6) = 0$
Hence, (2x $-$ 3) is a factor of the given polynomial.

Page No 91:

Answer:

$p (x) = 3 x^{3} + x^{2} - 20 x + 12$

$g (x) = 3 x - 2 = 3 (x - \frac{2}{3})$

Putting $x = \frac{2}{3}$ in p(x), we get

$p (\frac{2}{3}) = 3 \times {(\frac{2}{3})}^{3} + {(\frac{2}{3})}^{2} - 20 \times \frac{2}{3} + 12 = \frac{8}{9} + \frac{4}{9} - \frac{40}{3} + 12$ $= \frac{8 + 4 - 120 + 108}{9} = \frac{120 - 120}{9} = 0$

Therefore, by factor theorem, (3x − 2) is a factor of p(x).

Hence, g(x) is a factor of p(x).

Page No 91:

Answer:

Let:
$p (x) = 7 x^{2} - 4 \sqrt{2} x - 6$
Here,
$x - \sqrt{2} = 0 \Rightarrow x = \sqrt{2}$
By the factor theorem, $(x - \sqrt{2})$ is a factor of the given polynomial if $p (\sqrt{2}) = 0$
Thus, we have:
$p (\sqrt{2}) = [7 \times {(\sqrt{2})}^{2} - 4 \sqrt{2} \times \sqrt{2} - 6] = (14 - 8 - 6) = 0$
Hence, $(x - \sqrt{2})$ is a factor of the given polynomial.

Page No 91:

Answer:

Let:
$p (x) = 2 \sqrt{2} x^{2} + 5 x + \sqrt{2}$
Here,
$x + \sqrt{2} = 0 \Rightarrow x = - \sqrt{2}$
By the factor theorem, $(x + \sqrt{2})$ will be a factor of the given polynomial if $p (- \sqrt{2})$ = 0.
Thus, we have:
$p (- \sqrt{2}) = [2 \sqrt{2} \times {(- \sqrt{2})}^{2} + 5 \times (- \sqrt{2}) + \sqrt{2}] = (4 \sqrt{2} - 5 \sqrt{2} + \sqrt{2}) = 0$
Hence, $(x + \sqrt{2})$ is a factor of the given polynomial.

Page No 91:

Answer:

Let f(p) = p¹⁰ – 1 and g(p) = p¹¹ – 1.

Putting p = 1 in f(p), we get

f(1) = 1¹⁰ − 1 = 1 − 1 = 0

Therefore, by factor theorem, (p – 1) is a factor of (p¹⁰ – 1).

Now, putting p = 1 in g(p), we get

g(1) = 1¹¹ − 1 = 1 − 1 = 0

Therefore, by factor theorem, (p – 1) is a factor of (p¹¹ – 1).

Page No 91:

Answer:

Let:
$f (x) = 2 x^{3} + 9 x^{2} + x + k$
$(x - 1) is a factor of f (x) = 2 x^{3} + 9 x^{2} + x + k . \Rightarrow f (1) = 0 \Rightarrow 2 \times 1^{3} + 9 \times 1^{2} + 1 + k = 0 \Rightarrow 12 + k = 0 \Rightarrow k = - 12$
Hence, the required value of k is $-$ 12.

Page No 91:

Answer:

Let:
$f (x) = 2 x^{3} - 3 x^{2} - 18 x + a$
$(x - 4) is a factor of f (x) = 2 x^{3} - 3 x^{2} - 18 x + a . \Rightarrow f (4) = 0 \Rightarrow 2 \times 4^{3} - 3 \times 4^{2} - 18 \times 4 + a = 0 \Rightarrow 8 + a = 0 \Rightarrow a = - 8$
Hence, the required value of a is $-$ 8.

Page No 91:

Answer:

Let f(x) = ax³+ x² – 2x + 4a – 9

It is given that (x + 1) is a factor of f(x).

Using factor theorem, we have

f(−1) = 0
$\Rightarrow a \times {(- 1)}^{3} + {(- 1)}^{2} - 2 \times (- 1) + 4 a - 9 = 0 \Rightarrow - a + 1 + 2 + 4 a - 9 = 0 \Rightarrow 3 a - 6 = 0 \Rightarrow 3 a = 6 \Rightarrow a = 2$
Thus, the value of a is 2.

Page No 91:

Answer:

Let f(x) = x⁵– 4a² x³ + 2x + 2a +3

It is given that (x + 2a) is a factor of f(x).

Using factor theorem, we have

f(−2a) = 0
$\Rightarrow {(- 2 a)}^{5} - 4 a^{2} \times {(- 2 a)}^{3} + 2 \times (- 2 a) + 2 a + 3 = 0 \Rightarrow - 32 a^{5} - 4 a^{2} \times (- 8 a^{3}) + 2 \times (- 2 a) + 2 a + 3 = 0 \Rightarrow - 32 a^{5} + 32 a^{5} - 4 a + 2 a + 3 = 0 \Rightarrow - 2 a + 3 = 0$
$\Rightarrow 2 a = 3 \Rightarrow a = \frac{3}{2}$
Thus, the value of a is $\frac{3}{2}$ .

Page No 91:

Answer:

Let f(x) = 8x⁴+ 4x³ – 16x² + 10x + m

It is given that $(2 x - 1) = 2 (x - \frac{1}{2})$ is a factor of f(x).

Using factor theorem, we have

$f (\frac{1}{2}) = 0 \Rightarrow 8 \times {(\frac{1}{2})}^{4} + 4 \times {(\frac{1}{2})}^{3} - 16 \times {(\frac{1}{2})}^{2} + 10 \times \frac{1}{2} + m = 0 \Rightarrow \frac{1}{2} + \frac{1}{2} - 4 + 5 + m = 0$
$\Rightarrow 2 + m = 0 \Rightarrow m = - 2$
Thus, the value of m is −2.

Page No 91:

Answer:

Let:
$f (x) = x^{4} - x^{3} - 11 x^{2} - x + a$
Now,
$x + 3 = 0 \Rightarrow x = - 3$
By the factor theorem, $f (x) is exactly divisible by (x + 3) if f (- 3) = 0$ .
Thus, we have:
$f (- 3) = [{(- 3)}^{4} - {(- 3)}^{3} - 11 \times {(- 3)}^{2} - (- 3) + a] = (81 + 27 - 99 + 3 + a) = 12 + a$
Also,
$f (- 3) = 0 \Rightarrow 12 + a = 0 \Rightarrow a = - 12$
Hence, $f (x) is exactly divisible by (x + 3) when a is - 12 .$

Page No 91:

Answer:

Let:
$f (x) = x^{3} - 3 x^{2} - 13 x + 15$
And,
$g (x) = x^{2} + 2 x - 3$
$= x^{2} + x - 3 x - 3 = x (x - 1) + 3 (x - 1) = (x - 1) (x + 3)$
Now, $f (x) will be exactly divisible by g (x) if it is exactly divisible by (x - 1) as well as (x + 3)$ .
For this, we must have:
$f (1) = 0 and f (- 3) = 0$
Thus, we have:
$f (1) = (1^{3} - 3 \times 1^{2} - 13 \times 1 + 15) = (1 - 3 - 13 + 15) = 0$
And,
$f (- 3) = [{(- 3)}^{3} - 3 \times {(- 3)}^{2} - 13 \times (- 3) + 15] = (- 27 - 27 + 39 + 15) = 0$
$f (x) is exactly divisible by (x - 1) as well as (x + 3)$ . So, $f (x) is exactly divisible by (x - 1) (x + 3)$ .
Hence, $f (x) is exactly divisible b y x^{2} + 2 x - 3$ .

Page No 91:

Answer:

Let:
$f (x) = x^{3} + a x^{2} + b x + 6$
$(x - 2) is a factor of f (x) = x^{3} + a x^{2} + b x + 6 . \Rightarrow f (2) = 0 \Rightarrow 2^{3} + a \times 2^{2} + b \times 2 + 6 = 0 \Rightarrow 14 + 4 a + 2 b = 0 \Rightarrow 4 a + 2 b = - 14 \Rightarrow 2 a + b = - 7 . . . (1)$
Now,
$x - 3 = 0 \Rightarrow x = 3$
By the factor theorem, we can say:
$When f (x) will be divided by (x - 3), 3 will be its remainder . \Rightarrow f (3) = 3$

Now,
$f (3) = 3^{3} + a \times 3^{2} + b \times 3 + 6 = (27 + 9 a + 3 b + 6) = 33 + 9 a + 3 b$
Thus, we have:
$f (3) = 3 \Rightarrow 33 + 9 a + 3 b = 3 \Rightarrow 9 a + 3 b = - 30 \Rightarrow 3 a + b = - 10 . . . (2)$
Subtracting (1) from (2), we get:
a = $-$ 3
By putting the value of a in (1), we get the value of b, i.e., $-$ 1.
∴ a = $-$ 3 and b = $-$ 1

Page No 91:

Answer:

Let:
$f (x) = x^{3} - 10 x^{2} + a x + b$
Now,
$x - 1 = 0 \Rightarrow x = 1$
By the factor theorem, we can say:
$f (x) will be exactly divisible by (x - 1) if f (1) = 0$ .
Thus, we have:
$f (1) = 1^{3} - 10 \times 1^{2} + a \times 1 + b = (1 - 10 + a + b) = - 9 + a + b$
∴ $f (1) = 0 \Rightarrow a + b = 9 . . . (1)$
Also,
$x - 2 = 0 \Rightarrow x = 2$
By the factor theorem, we can say:
$f (x) will be exactly divisible by (x - 2) if f (2) = 0$ .
Thus, we have:
$f (2) = 2^{3} - 10 \times 2^{2} + a \times 2 + b = (8 - 40 + 2 a + b) = - 32 + 2 a + b$
∴ $f (2) = 0 \Rightarrow 2 a + b = 32 . . . (2)$

$Subtracting (1) from (2), we get : a = 23$
Putting the value of a, we get the value of b, i.e., $-$ 14.
∴ a = 23 and b = $-$ 14

Page No 91:

Answer:

Let:
$f (x) = x^{4} + a x^{3} - 7 x^{2} - 8 x + b$
Now,
$x + 2 = 0 \Rightarrow x = - 2$
By the factor theorem, we can say:
$f (x) will be exactly divisible by (x + 2) if f (- 2) = 0$ .
Thus, we have:
$f (- 2) = [{(- 2)}^{4} + a \times {(- 2)}^{3} - 7 \times {(- 2)}^{2} - 8 \times (- 2) + b] = (16 - 8 a - 28 + 16 + b) = (4 - 8 a + b)$
∴ $f (- 2) = 0 \Rightarrow 8 a - b = 4 . . . (1)$
Also,
$x + 3 = 0 \Rightarrow x = - 3$
By the factor theorem, we can say:
$f (x) will be exactly divisible by (x + 3) if f (- 3) = 0$ .
Thus, we have:
$f (- 3) = [{(- 3)}^{4} + a \times {(- 3)}^{3} - 7 \times {(- 3)}^{2} - 8 \times (- 3) + b] = (81 - 27 a - 63 + 24 + b) = (42 - 27 a + b)$
∴ $f (- 3) = 0 \Rightarrow 27 a - b = 42 . . . (2)$
$Subtracting (1) from (2), we get : \Rightarrow 19 a = 38 \Rightarrow a = 2$
Putting the value of a, we get the value of b, i.e., 12.
∴ a = 2 and b = 12

Page No 91:

Answer:

Let f(x) = px² + 5x + r

It is given that (x – 2) is a factor of f(x).

Using factor theorem, we have

$f (2) = 0 \Rightarrow p \times 2^{2} + 5 \times 2 + r = 0 \Rightarrow 4 p + r = - 10 . . . . . (1)$
Also, $(x - \frac{1}{2})$ is a factor of f(x).

Using factor theorem, we have

$f (\frac{1}{2}) = 0 \Rightarrow p \times {(\frac{1}{2})}^{2} + 5 \times \frac{1}{2} + r = 0 \Rightarrow \frac{p}{4} + r = - \frac{5}{2} \Rightarrow p + 4 r = - 10 . . . . . (2)$
From (1) and (2), we have

$4 p + r = p + 4 r \Rightarrow 4 p - p = 4 r - r \Rightarrow 3 p = 3 r \Rightarrow p = r$

Page No 91:

Answer:

Let f(x) = 2x⁴ – 5x³ + 2x² – x + 2 and g(x) = x² – 3x + 2.

$x^{2} - 3 x + 2 = x^{2} - 2 x - x + 2 = x (x - 2) - 1 (x - 2) = (x - 1) (x - 2)$
Now, f(x) will be divisible by g(x) if f(x) is exactly divisible by both (x − 1) and (x − 2).

Putting x = 1 in f(x), we get

f(1) = 2 × 1⁴ – 5 × 1³ + 2 × 1² – 1 + 2 = 2 – 5 + 2 – 1 + 2 = 0

By factor theorem, (x − 1) is a factor of f(x). So, f(x) is exactly divisible by (x − 1).

Putting x = 2 in f(x), we get

f(2) = 2 × 2⁴ – 5 × 2³ + 2 × 2² – 2 + 2 = 32 – 40 + 8 – 2 + 2 = 0

By factor theorem, (x − 2) is a factor of f(x). So, f(x) is exactly divisible by (x − 2).

Thus, f(x) is exactly divisible by both (x − 1) and (x − 2).

Hence, f(x) = 2x⁴ – 5x³ + 2x² – x + 2 is exactly divisible by (x − 1)(x − 2) = x² – 3x + 2.

Page No 91:

Answer:

Let k be added to 2x⁴– 5x³+ 2x² – x – 3 so that the result is exactly divisible by (x – 2). Here, k is a constant.

∴ f(x) = 2x⁴– 5x³+ 2x² – x – 3 + k is exactly divisible by (x – 2).

Using factor theorem, we have

$f (2) = 0 \Rightarrow 2 \times 2^{4} - 5 \times 2^{3} + 2 \times 2^{2} - 2 - 3 + k = 0 \Rightarrow 32 - 40 + 8 - 5 + k = 0 \Rightarrow - 5 + k = 0 \Rightarrow k = 5$
Thus, 5 must be added to 2x⁴– 5x³+ 2x² – x – 3 so that the result is exactly divisible by (x – 2).

Page No 91:

Answer:

Dividing (x⁴+ 2x³– 2x² + 4x + 6) by (x²+ 2x – 3) using long division method, we have

Here, the remainder obtained is (2x + 9).

Thus, the remainder (2x + 9) must be subtracted from (x⁴+ 2x³– 2x² + 4x + 6) so that the result is exactly divisible by (x²+ 2x – 3).

Page No 91:

Answer:

Let f(x) = xⁿ+ aⁿ

Putting x = −a in f(x), we get

f(−a) = (−a)ⁿ+ aⁿ

If n is any odd positive integer, then

f(−a) = (−a)ⁿ+ aⁿ = −aⁿ+ aⁿ = 0

Therefore, by factor theorem, (x + a) is a factor of (xⁿ+ aⁿ) for any odd positive integer.

Page No 92:

Answer:

(c) $\sqrt{2} x^{2} - \sqrt{3} x + 6$

Clearly, $\sqrt{2} x^{2} - \sqrt{3} x + 6$ is a polynomial in one variable because it has only non-negative integral powers of x.

Page No 92:

Answer:

(d) $x^{2} + \frac{2 x^{3 / 2}}{\sqrt{x}} + 6$

We have:

$x^{2} + \frac{2 x^{\frac{3}{2}}}{\sqrt{x}} + 6 = x^{2} + 2 x^{\frac{3}{2}} x^{- \frac{1}{2}} + 6$
$= x^{2} + 2 x + 6$

It a polynomial because it has only non-negative integral powers of x.

Page No 93:

Answer:

Page No 93:

Answer:

(d) −4

$-$ 4 is a constant polynomial of degree zero.

Page No 93:

Answer:

(d) 0

0 is a polynomial whose degree is not defined.

Page No 93:

Answer:

(d) x² + 5x + 4

$x^{2} + 5 x + 4$ is a polynomial of degree 2. So, it is a quadratic polynomial.

Page No 93:

Answer:

(b) x + 1

Clearly, $x + 1$ is a polynomial of degree 1. So, it is a linear polynomial.

Page No 93:

Answer:

(b) x² + 4

Clearly, $x^{2} + 4$ is an expression having two non-zero terms. So, it is a binomial.

Page No 93:

Answer:

(d) 0

$\sqrt{3}$ is a constant term, so it is a polynomial of degree 0.

Page No 93:

Answer:

Page No 93:

Answer:

(d) not defined

Zero of the zero polynomial is not defined.

Page No 93:

Answer:

(d) 8

Let:
$p (x) = (x + 4)$

$∴ p (- x) = (- x) + 4 = - x + 4$
Thus, we have:
$p (x) + p (- x) = \{(x + 4) + (- x + 4)\}$
= 4 + 4
=8

Page No 93:

Answer:

(b) 1

$p (x) = x^{2} - 2 \sqrt{2} x + 1 ∴ p (2 \sqrt{2}) = {(2 \sqrt{2})}^{2} - 2 \sqrt{2} \times (2 \sqrt{2}) + 1$
= 8 $-$ 8 + 1
= 1

Page No 93:

Answer:

p(x) = 5x – 4x² + 3

Putting x = –1 in p(x), we get

p(–1) = 5 × (–1) – 4 × (–1)² + 3 = –5 – 4 + 3 = –6

Hence, the correct answer is option (d).

Page No 93:

Answer:

Let f(x) = x⁵¹ + 51

By remainder theorem, when f(x) is divided by (x + 1), then the remainder = f(−1).

Putting x = −1 in f(x), we get

f(−1) = (−1)⁵¹ + 51 = −1 + 51 = 50

∴ Remainder = 50

Thus, the remainder when (x⁵¹ + 51) is divided by (x + 1) is 50.

Hence, the correct answer is option (d).

Page No 93:

Answer:

(c) 2

$(x + 1) is a factor of 2 x^{2} + k x . So, - 1 is a zero of 2 x^{2} + k x . Thus, we have : 2 \times {(- 1)}^{2} + k \times (- 1) = 0 \Rightarrow 2 - k = 0 \Rightarrow k = 2$

Page No 93:

Answer:

(d) 21

$x - 2 = 0 \Rightarrow x = 2$
By the remainder theorem, we know that when p(x) is divided by (x $-$ 2), the remainder is p(2).
Thus, we have:
$p (2) = 2^{4} + 2 \times 2^{3} - 3 \times 2^{2} + 2 - 1 = 16 + 16 - 12 + 1 = 21$

Page No 94:

Answer:

(d) 4

$x + 2 = 0 \Rightarrow x = - 2$
By the remainder theorem, we know that when p(x) is divided by (x + 2), the remainder is p( $-$ 2).
Now, we have:
$p (- 2) = {(- 2)}^{3} - 3 \times {(- 2)}^{2} + 4 \times (- 2) + 32 = - 8 - 12 - 8 + 32 = 4$

Page No 94:

Answer:

(c) −2

$2 x - 1 = 0 \Rightarrow x = \frac{1}{2}$
By the remainder theorem, we know that when p(x) is divided by (2x $-$ 1), the remainder is $p (\frac{1}{2})$ .
Now, we have:
$p (\frac{1}{2}) = 4 \times {(\frac{1}{2})}^{3} - 12 \times {(\frac{1}{2})}^{2} + 11 \times \frac{1}{2} - 5 = \frac{1}{2} - 3 + \frac{11}{2} - 5 = - 2$

Page No 94:

Answer:

By remainder theorem, when p(x) = x³– ax²+ x is divided by (x – a), then the remainder = p(a).

Putting x = a in p(x), we get

p(a) = a³– a × a²+ a = a³– a³+ a = a

∴ Remainder = a

Hence, the correct answer is option (b).

Page No 94:

Answer:

(c) −a

$x + a = 0 \Rightarrow x = - a$
By the remainder theorem, we know that when p (x) is divided by (x + a), the remainder is p (−a).
Thus, we have:
$p (- a) = {(- a)}^{3} + a \times {(- a)}^{2} + 2 \times (- a) + a = - a^{3} + a^{3} - 2 a + a = - a$

Page No 94:

Answer:

(b) x³ − 2x² − x − 2

Let:
f(x) = x³ − 2x² − x − 2
By the factor theorem, (x + 1) will be a factor of f(x) if f ( $-$ 1) = 0.
We have:
$f (- 1) = {(- 1)}^{3} + 2 \times {(- 1)}^{2} - (- 1) - 2 = - 1 + 2 + 1 - 2 = 0$
Hence, (x + 1) is a factor of $f (x) = x^{3} + 2 x^{2} - x - 2$ .

Page No 94:

Answer:

The zero of the polynomial p(x) can be obtained by putting p(x) = 0.

$p (x) = 0 \Rightarrow 2 x + 5 = 0 \Rightarrow 2 x = - 5 \Rightarrow x = - \frac{5}{2}$
Hence, the correct answer is option (b).

Page No 94:

Answer:

The given polynomial is p(x) = x² + x – 6.

Putting x = 2 in p(x), we get

p(2) = 2² + 2 – 6 = 4 + 2 – 6 = 0

Therefore, x = 2 is a zero of the polynomial p(x).

Putting x = –3 in p(x), we get

p(–3) = (–3)² – 3 – 6 = 9 – 9 = 0

Therefore, x = –3 is a zero of the polynomial p(x).

Thus, 2 and –3 are the zeroes of the given polynomial p(x).

Hence, the correct answer is option (c).

Page No 94:

Answer:

The given polynomial is p(x) = 2x² + 5x – 3.

Putting $x = \frac{1}{2}$ in p(x), we get

$p (\frac{1}{2}) = 2 \times {(\frac{1}{2})}^{2} + 5 \times \frac{1}{2} - 3 = \frac{1}{2} + \frac{5}{2} - 3 = 3 - 3 = 0$

Therefore, $x = \frac{1}{2}$ is a zero of the polynomial p(x).

Putting x = –3 in p(x), we get

$p (- 3) = 2 \times {(- 3)}^{2} + 5 \times (- 3) - 3 = 18 - 15 - 3 = 0$

Therefore, x = –3 is a zero of the polynomial p(x).

Thus, $\frac{1}{2}$ and –3 are the zeroes of the given polynomial p(x).

Hence, the correct answer is option (b).

Page No 94:

Answer:

The given polynomial is p(x) = 2x² + 7x – 4.

Putting $x = \frac{1}{2}$ in p(x), we get

$p (\frac{1}{2}) = 2 \times {(\frac{1}{2})}^{2} + 7 \times \frac{1}{2} - 4 = \frac{1}{2} + \frac{7}{2} - 4 = 4 - 4 = 0$

Therefore, $x = \frac{1}{2}$ is a zero of the polynomial p(x).

Putting x = –4 in p(x), we get

$p (- 4) = 2 \times {(- 4)}^{2} + 7 \times (- 4) - 4 = 32 - 28 - 4 = 32 - 32 = 0$

Therefore, x = –4 is a zero of the polynomial p(x).

Thus, $\frac{1}{2}$ and –4 are the zeroes of the given polynomial p(x).

Hence, the correct answer is option (c).

Page No 94:

Answer:

(b) 5

$(x + 5) is a factor of p (x) = x^{3} - 20 x + 5 k . ∴ p (- 5) = 0 \Rightarrow {(- 5)}^{3} - 20 \times (- 5) + 5 k = 0 \Rightarrow - 125 + 100 + 5 k = 0 \Rightarrow 5 k = 25 \Rightarrow k = 5$

Page No 94:

Answer:

(b) m = 7, n = −18

Let:
$p (x) = x^{3} + 10 x^{2} + m x + n$
Now,
$x + 2 = 0 \Rightarrow x = - 2$
(x + 2) is a factor of p(x).
So, we have p( $-$ 2)=0
$\Rightarrow {(- 2)}^{3} + 10 \times {(- 2)}^{2} + m \times (- 2) + n = 0 \Rightarrow - 8 + 40 - 2 m + n = 0 \Rightarrow 32 - 2 m + n = 0 \Rightarrow 2 m - n = 32 . . . . . (i)$
Now,
$x - 1 = 0 \Rightarrow x = 1$
Also,
(x $-$ 1) is a factor of p(x).
We have:
p(1) = 0
$\Rightarrow 1^{3} + 10 \times 1^{2} + m \times 1 + n = 0 \Rightarrow 1 + 10 + m + n = 0 \Rightarrow 11 + m + n = 0 \Rightarrow m + n = - 11 . . . . . (ii) From (i) and (ii), we get : 3 m = 21 \Rightarrow m = 7$
By substituting the value of m in (i), we get n = −18.
∴ m = 7 and n = −18

Page No 94:

Answer:

(a) 1

Let:
$p (x) = x^{100} + 2 x^{99} + k$
Now,
$x + 1 = 0 \Rightarrow x = - 1$
$(x + 1) is divisible by p (x) . ∴ p (- 1) = 0 \Rightarrow {(- 1)}^{100} + 2 \times {(- 1)}^{99} + k = 0 \Rightarrow 1 - 2 + k = 0 \Rightarrow - 1 + k = 0 \Rightarrow k = 1$

Page No 94:

Answer:

(d) −3

Let:
$p (x) = 2 x^{3} - k x^{2} + 3 x + 10$
Now,
$x + 2 = 0 \Rightarrow x = - 2$
$p (x) is completely divisible by (x + 2) . ∴ p (- 2) = 0 \Rightarrow 2 \times {(- 2)}^{3} - k \times {(- 2)}^{2} + 3 \times (- 2) + 10 = 0 \Rightarrow - 16 - 4 k - 6 + 10 = 0 \Rightarrow - 12 - 4 k = 0 \Rightarrow 4 k = - 12 \Rightarrow k = \frac{- 12}{4} \Rightarrow k = - 3$

Page No 94:

Answer:

(b) 0, 3

Let:
$p (x) = x^{2} - 3 x$
Now, we have:
$p (x) = 0 \Rightarrow x^{2} - 3 x = 0$
$\Rightarrow x (x - 3) = 0 \Rightarrow x = 0 and x - 3 = 0 \Rightarrow x = 0 and x = 3$

Page No 95:

Answer:

(d) $\frac{1}{\sqrt{3}} and \frac{- 1}{\sqrt{3}}$

Let:
$p (x) = 3 x^{2} - 1$

$To find the zeroes of p (x), we have : p (x) = 0 \Rightarrow 3 x^{2} - 1 = 0$
$\Rightarrow 3 x^{2} = 1 \Rightarrow x^{2} = \frac{1}{3} \Rightarrow x = \pm \frac{1}{\sqrt{3}} \Rightarrow x = \frac{1}{\sqrt{3}} and x = - \frac{1}{\sqrt{3}}$
Hence, the correct answer is option D.

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