R.d Sharma 2022 _mcqs for Class 9 Maths Chapter 5 - Factorization Of Algebraic Expressions

Answer:

The given expression to be factorized is

Take common from the first two terms and from the last two terms. That is

Finally, take commonfrom the two terms. That is

So, the correct choice is (d).

Answer:

The given expression to be factorized is

This can be written in the form

Recall the formula

Using the above formula, we have 

So, the correct choice is (a).

Answer:

The given expression to be factorized is

This can be written in the form

Recall the formula

Using the above formula, we have 

So, the correct choice is (c).

Answer:

The given expression to be factorized is

Recall the formula for difference of two cubes

Using the above formula, we have,

So, the correct choice is (a).

Answer:

The given expression is

Let, and. Then the given expression becomes

Note that:

Recall the formula

When, this becomes

So, we have the new formula

, when.

Using the above formula, the value of the given expression is

So, the correct choice is (b).

Answer:

The given expression is

This can be written in the form

Assumeand. Then the given expression can be rewritten as

Recall the formula for difference of two cubes

Using the above formula, the expression becomes

Note that both a and b are positive, unequal. So, neithernor any factor of it can be zero.

Therefore we can cancel the termfrom both numerator and denominator. Then the expression becomes

So, the correct choice is (a).

Answer:

The given expression is

Assumeand. Then the given expression can be rewritten as

Recall the formula for sum of two cubes

Using the above formula, the expression becomes

Note that both and b are positive. So, neithernor any factor of it can be zero.

Therefore we can cancel the termfrom both numerator and denominator. Then the expression becomes

So, the correct choice is (b).

Answer:

The given expression to be factorized is

Take commonfrom the last three terms and then we have

So, the correct choice is (c).

Answer:

The given expression to be factorized is

This can be written in the form

So, the correct choice is (a).

Answer:

The given expression to be factorized is

This can be arrange in the form

Let. Then the above expression becomes

Put.

So, the correct choice is (a).

Answer:

The given expression to be factorized is

This can be written in the form

Take common x from the first two terms andfrom the last two terms. Then we have

Finally, take commonfrom the above expression,

So, the correct choice is (d).

Answer:

Given: $\frac{x}{y}+\frac{y}{x}=-1(x, y\ne 0)$             .....(1)
$$\begin{gathered} \Rightarrow \frac{x^2+y^2}{yx}=-1 \\ \Rightarrow x^2+y^2=-yx \\ \Rightarrow x^2+yx+y^2=0 .....\left(2\right) \end{gathered}$$

Now,

$$\begin{gathered} x^3-y^3=\left(x-y\right)\left(x^2+xy+y^2\right) \left[\because a^3-b^3=\left(a-b\right)\left(a^2+ab+b^2\right)\right] \\ \Rightarrow x^3-y^3=\left(x-y\right)·0 \left[\mathrm{From} \left(1\right)\right] \\ \Rightarrow x^3-y^3=0 \end{gathered}$$

Hence, the correct answer is option (c).

Answer:

Disclaimer: In the question, y³ is incorrectly written as y² and the calculations are shown accordingly.

(x + y)³ – (x³ + y³)
$$\begin{gathered} =x^3+y^3+3xy\left(x+y\right)-\left(x^3+y^3\right) \\ =3xy\left(x+y\right) \end{gathered}$$

Thus, there are two factors 3xy and (x + y).

Hence, the correct answer is option (d).

Answer:

Statement-2 (Reason): If a + b + c = 0 then a³ + b³ + c³ = 3abc

$$\begin{gathered} \Rightarrow a^3+b^3+c^3–3abc=\left(a+b+c\right)\left(a^2+b^{2 }+c^{2 }–ab–bc–ca\right) \\ \Rightarrow a^3+b^3+c^3–3abc=0·\left(a^2+b^{2 }+c^{2 }–ab–bc–ca\right) \\ \Rightarrow a^3+b^3+c^3–3abc=0 \\ \Rightarrow a^3+b^3+c^3=3abc \end{gathered}$$

Thus, Statement-2 is true.

Statement-1 (Assertion): (a – b)³ + (b – c)³ + (c – a)³ = 3(a – b) (b – c) (c – a)

Here, a – b + b – c + c – a = 0
Now, According to the Statement-2
${(a-b)}^3+{(b-c)}^3+{(c-a)}^3=3(a-b)(b-c)(c-a)$

Thus, Statement-1 is true.
So, Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1.
 
Hence, the correct answer is option (a).

Answer:

Statement-2 (Reason): If a + b + c = 0 then a³ + b³ + c³ = 3abc

$$\begin{gathered} \Rightarrow a^3+b^3+c^3–3abc=\left(a+b+c\right)\left(a^2+b^{2 }+c^{2 }–ab–bc–ca\right) \\ \Rightarrow a^3+b^3+c^3–3abc=0·\left(a^2+b^{2 }+c^{2 }–ab–bc–ca\right) \\ \Rightarrow a^3+b^3+c^3–3abc=0 \\ \Rightarrow a^3+b^3+c^3=3abc \end{gathered}$$

Thus, Statement-2 is true.

Statement-1 (Assertion): If 3x = a + b + c, then (x – a)³ + (x – b)³ + (x – c)³ = 3(x – a) (x – b) (x – c)

Given that, 3x = a + b + c
Here,
 $\begin{array}{rcl}(x-a)+(x-b)+(x-c)& =& 3x-a-b-c\\ & =& 3x-\left(a+b+c\right)\\ & =& 3x-3x \left(\mathrm{Given}\right)\\ & =& 0 \\ & & \\ & & \end{array}$

Now, According to the Statement-2
${(x-a)}^3+{(x-b)}^3+{(x-c)}^3=3(x-a)(x-b)(x-c)$   
Thus, Statement-1 is true.
So, Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1.
 
Hence, the correct answer is option (a).

Answer:

Statement-2 (Reason): a³ + b³ + c³ – 3abc = (a + b + c) {(a + b + c)² – 3(ab + bc + ca)}

Since
$$\begin{gathered} {\left(a+b+c\right)}^2=a^2+b^2+c^2+2\left(ab+ac+bc\right) \\ \Rightarrow a^2+b^2+c^2={\left(a+b+c\right)}^2-2\left(ab+ac+bc\right) .....\left(1\right) \end{gathered}$$

Now,
$$\begin{gathered} a^3+b^3+c^3–3abc=\left(a+b+c\right)\left(a^2+b^{2 }+c^{2 }–ab–bc–ca\right) \\ \Rightarrow a^3+b^3+c^3–3abc=\left(a+b+c\right)\left[{\left(a+b+c\right)}^2-2\left(ab+ac+bc\right){ }^2–\left(ab+bc+ca\right)\right] \\ \Rightarrow a^3+b^3+c^3–3abc=\left(a+b+c\right)\left[{\left(a+b+c\right)}^2-3\left(ab+ac+bc\right)\right] \end{gathered}$$

Thus, Statement-2 is true.

Statement-1 (Assertion): If a + b + c = 5 and ab + bc + ca = 10, then a³ + b³ + c³ – 3abc = 25

Given that, a + b + c = 5 and ab + bc + ca = 10.
$$\begin{gathered} a^3+b^3+c^3–3abc=\left(a+b+c\right)\left[{\left(a+b+c\right)}^2-3\left(ab+ac+bc\right)\right] \\ \Rightarrow a^3+b^3+c^3–3abc=\left(5\right)\left[{\left(5\right)}^2-3\left(10\right)\right] \\ \Rightarrow a^3+b^3+c^3–3abc=\left(5\right)\left[25-30\right] \\ \Rightarrow a^3+b^3+c^3–3abc=\left(5\right)\left[-5\right] \\ \Rightarrow a^3+b^3+c^3–3abc=-25 \end{gathered}$$

Thus, Statement-1 is false.
So, Statement-1 is false, Statement-2 is true.

Hence, the correct answer is option (d).

Answer:

Statement-2 (Reason): If a + b + c = 9 and a² + b² + c² = 35, then ab + bc + ca = 23
Given that, a + b + c = 9 and a² + b² + c² = 35.

$$\begin{gathered} {\left(a+b+c\right)}^2=a^2+b^2+c^2+2\left(ab+ac+bc\right) \\ \Rightarrow {\left(9\right)}^2=35+2\left(ab+ac+bc\right) \left(\mathrm{Given}\right) \\ \Rightarrow 81=35+2\left(ab+ac+bc\right) \\ \Rightarrow 46=2\left(ab+ac+bc\right) \\ \Rightarrow 23=\left(ab+ac+bc\right) \end{gathered}$$

Thus, Statement-2 is true.

Statement-1 (Assertion): If a, b, c are all non-zero such that a + b + c = 0, then $\frac{a^2}{bc}+\frac{b^2}{ca}+\frac{c^2}{ab}=3$

Given that, a + b + c = 0
As we know that, if a + b + c = 0, then a³ + b³ + c³ = 3abc            .....(1)

$\begin{array}{rcl}\frac{a^2}{bc}+\frac{b^2}{ca}+\frac{c^2}{ab}& =& \frac{a^3+b^3+c^3}{abc}\\ & =& \frac{3abc}{abc} \left[\mathrm{From} \left(1\right)\right]\\ & =& 3\end{array}$

Thus, Statement-1 is true.
So, Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1.
 
Hence, the correct answer is option (b).

Answer:

The given expression to be factorized is

This can be written in the form 

So, the correct choice is (a).

Answer:

The given expression is

Recall the formula

Using the above formula the given expression becomes

Given that

Therefore the value of the given expression is

So, the correct choice is (c).

Answer:

The given equation is

Recall the formula 

Using the above formula, we have

, provided.

So, the correct choice is (d).

Answer:

The given equation is

x³ − 3x² + 3x − 7 = (x + 1) (ax² + bx + c)

This can be written as

$$\begin{gathered} x^3-3x^2+3x-7=\left(x+1\right)\left(a{x}^2+bx+c\right) \\ \Rightarrow x^3-3x^2+3x-7=a{x}^3+b{x}^2+cx+a{x}^2+bx+c \\ \Rightarrow x^3-3x^2+3x-7=a{x}^3+\left(a+b\right)x^2+\left(b+c\right)x+c \end{gathered}$$
 

Comparing the coefficients on both sides of the equation.

We get,

c = -7 .......(4)

Putting the value of a from (1) in (2)

We get,

So the value of a, b and c is 1, – 4 and -7 respectively.

Therefore,

a + b + c =1 - 4 - 7 = -10

So, the correct choice is (c).

Answer:

Statement-2 (Reason): a³ – b³ = (a – b)(a² – ab + b²)

$\begin{array}{rcl}\left(a-b\right)\left(a^2-ab+b^2\right)& =& a^3-a^{2}b+a{b}^2-a^{2}b+a{b}^2-b^3\\ & =& a^3-a^{2}b-a^{2}b+a{b}^2+a{b}^2-b^3\\ & =& a^3-2a^{2}b+2a{b}^2-b^3\end{array}$

Thus, Statement-2 is false.

Statement-1 (Assertion): The value of $\frac{{(0.027)}^3 + {(0.023)}^3}{{(0.027)}^2 - (0.027) (0.023) + {(0.023)}^2} \mathrm{is} 0.05$

$\begin{array}{rcl}\frac{{(0.027)}^3 + {(0.023)}^3}{{(0.027)}^2 - (0.027) (0.023) + {(0.023)}^2}& =& \frac{\left(0.027+0.023\right)\left[{\left(0.027\right)}^2+{\left(0.023\right)}^2-\left(0.027\times 0.023\right)\right]}{{(0.027)}^2 - (0.027) (0.023) + {(0.023)}^2} \left[\because a^3-b^3=\left(a+b\right)\left(a^2+b^2-ab\right)\right]\\ & =& \left(0.027+0.023\right)\\ & =& 0.05\end{array}$

Thus, Statement-1 is true.
So, Statement-1 is true, Statement-2 is false.
 
Hence, the correct answer is option (c).