RD Sharma 2020 solution class 9 chapter 5 Factorization of Algebraic Expressions MCQS

MCQS

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Question 1:

The factors of x³ −x²y − xy² + y³ are

(a)(x + y) (x² − xy + y²)

(b) (x + y) (x² + xy + y²)

(c) (x + y)² (x − y)

(d) (x − y)² (x + y)

Answer 1:

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).

Question 2:

The factors of x³ − 1 + y³ + 3xy are

(a) (x − 1 + y) (x² + 1 + y² + x + y − xy)

(b) (x + y + 1) (x² + y² + 1 −xy − x − y)

(c) (x − 1 + y) (x² − 1 − y² + x + y + xy)

(d) 3(x + y −1) (x² + y² − 1)

Answer 2:

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).

Question 3:

The factors of 8a³ + b³ − 6ab + 1 are

(a) (2a + b − 1) (4a² + b² + 1 − 3ab − 2a)

(b) (2a − b + 1) (4a² + b² − 4ab + 1 − 2a + b)

(c) (2a + b + 1) (4a² + b² + 1 −2ab − b − 2a)

(d) (2a − 1 + b) (4a² + 1 − 4a − b − 2ab)

Answer 3:

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).

Question 4:

(x + y)³ − (x − y)³ can be factorized as

(a) 2y (3x² + y²)

(b) 2x (3x² + y²)

(c) 2y (3y² + x²)

(d) 2x (x²+ 3y²) 

Answer 4:

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).

Question 5:

The expression (a − b)³ + (b − c)³ + (c −a)³ can be factorized as

(a) (a − b) (b − c) (c −a)

(b) 3(a − b) (b − c) (c −a)

(c) −3(a − b) (b −c) (c − a)

(d) (a + b + c) (a² + b² + c² − ab − bc − ca)

Answer 5:

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).

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Question 6:

The value of $\frac{(2.3{)}^3-0.027}{(2.3{)}^2+0.69+0.09}$

(a) 2

(b) 3

(c) 2.327

(d) 2.273

Answer 6:

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).

Question 7:

The value of $\frac{(0.013{)}^3+(0.007{)}^3}{(0.013{)}^2-0.013\times 0.007+(0.007{)}^2}$ is

(a) 0.006

(b) 0.02

(c) 0.0091

(d) 0.00185

Answer 7:

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).

Question 8:

Mark the correct alternative in each of the following:

The factors of a² − 1 − 2x − x² are

(a) (a − x + 1) (a − x − 1)

(b) (a + x − 1) (a − x + 1)

(c) (a + x +1) (a − x + 1)

(d) none of these

Answer 8:

The given expression to be factorized is

Take commonfrom the last three terms and then we have

So, the correct choice is (c).

Question 9:

The factors of x⁴ + x² + 25 are

(a) (x² + 3x + 5) (x² − 3x + 5)

(b) (x² + 3x + 5) (x² + 3x − 5)

(c) (x² + x +5) (x² − x + 5)

(d) none of these

Answer 9:

The given expression to be factorized is

This can be written in the form

So, the correct choice is (a).

Question 10:

The factors of x² + 4y² + 4y − 4xy − 2x − 8 are

(a) (x − 2y −4) (x − 2y + 2)

(b) (x − y + 2) (x − 4y − 4)

(c) (x + 2y − 4) (x + 2y + 2)

(d) none of these

Answer 10:

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).

Question 11:

The factors of x³ − 7x + 6 are

(a) x (x − 6) (x − 1)

(b) (x² − 6) (x − 1)

(c) (x + 1) (x + 2) (x + 3)

(d) (x − 1) (x + 3) (x − 2) 

Answer 11:

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).

Question 12:

The expression x⁴ + 4 can be factorized as

(a) (x² + 2x + 2) (x² − 2x + 2)

(b) (x² + 2x + 2) (x² + 2x + 2)

(c) (x² − 2x − 2) (x² − 2x + 2)

(d) (x² + 2) (x² − 2)

Answer 12:

The given expression to be factorized is

This can be written in the form 

So, the correct choice is (a).

Question 13:

If 3x = a + b + c, then the value of (x − a)³ + (x −b)³ + (x − c)³ − 3(x − a) (x − b) (x −c) is

(a) a + b + c

(b) (a − b) (b − c) (c − a)

(c) 0

(d) none of these

Answer 13:

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).

Question 14:

If (x + y)³ − (x − y)³ − 6y(x² − y²) = ky², then k =

(a) 1

(b) 2

(c) 4

(d) 8

Answer 14:

The given equation is

Recall the formula 

Using the above formula, we have

, provided.

So, the correct choice is (d).

Question 15:

If x³ − 3x² + 3x − 7 = (x + 1) (ax² + bx + c), then a + b + c =

(a) 4

(b) 12

(c) −10

(d) 3

Answer 15:

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).