RD Sharma 2020 solution class 9 chapter 6 Factorization of polynomial Expressions Exercise 6.4

Exercise 6.4

Page-6.24

Question 1:

In each of the following, use factor theorem to find whether polynomial g(x) is a factor of polynomial f(x) or, not: (1−7)

f(x) = x³ − 6x² + 11x − 6; g(x) = x − 3

Answer 1:

Given that:

By the factor theorem,

If g(x) is a factor of f(x)

i.e.

Then

As is zero therefore g(x), is the factor of polynomial f(x).

Question 2:

f(x) = 3x⁴ + 17x³ + 9x² − 7x − 10; g(x) = x + 5
 

Answer 2:

It is given that and 

By the factor theorem, g(x) is a factor of polynomial f(x)

i.e. 

Therefore,

$$\begin{gathered} f(-5)=3{\left(-5\right)}^4+17{\left(-5\right)}^3+9{\left(-5\right)}^2-7\left(-5\right)-10 \\ =3\times 625+17\times \left(-125\right)+225+35-10 \\ =1875-2125+250 \\ =0 \end{gathered}$$

Hence, g(x) is the factor of polynomial f(x).

Question 3:

f(x) = x⁵ + 3x⁴ − x³ − 3x² + 5x + 15, g(x) = x + 3

Answer 3:

It is given that and

By the factor theorem, g(x) is the factor of polynomial f(x).

i.e.

f (−3) = 0

Hence, g(x) is the factor of polynomial f (x).

Question 4:

f(x) = x³ −6x² − 19x + 84, g(x) = x − 7

Answer 4:

It is given that and

By the factor theorem, g(x) is the factor of polynomial f(x), if f (7) = 0.

Therefore, in order to prove that (x − 7) is a factor of f(x).

It is sufficient to show that f(7) = 0

Now,

Hence, (x − 7) is a factor of polynomial f(x).

Question 5:

f(x) = 3x³ + x² − 20x +12, g(x) = 3x − 2

Answer 5:

It is given that and

By the factor theorem,

(3x − 2) is the factor of f(x), if

Therefore,

In order to prove that (3x − 2) is a factor of f(x).

It is sufficient to show that

Now,

Hence, (3x − 2) is the factor of polynomial f(x).

Question 6:

f(x) = 2x³ − 9x² + x + 12, g(x) = 3 − 2x

Answer 6:

It is given that and

By factor theorem, (3 − 2x) is the factor of f(x), if = 0

Therefore,

In order to prove that (3 − 2x) is a factor of f(x). It is sufficient to show that

Now,

Hence, (3 − 2x), is the factor of polynomial f(x).

Question 7:

f(x) = x³ − 6x² + 11x − 6, g(x) = x² − 3x + 2

Answer 7:

It is given that and

We have

$\Rightarrow \left(x-2\right)$ and (x − 1) are factor of g(x) by the factor theorem.

To prove that (x − 2) and (x − 1) are the factor of f(x).

It is sufficient to show that f(2) and f(1) both are equal to zero.

And

Hence, g(x) is the factor of the polynomial f(x).

Question 8:

Show that (x − 2), (x + 3) and (x − 4) are factors of x³ − 3x² − 10x + 24.

Answer 8:

Let be the given polynomial.

By factor theorem,

and are the factor of f(x).

If and f(4) are all equal to zero.

Now,

also

And

Hence, and are the factor of polynomial f(x).

Question 9:

Show that (x + 4) , (x − 3) and (x − 7) are factors of x³ − 6x² − 19x + 84

Answer 9:

Let be the given polynomial.

By the factor theorem,

and are the factor of f(x).

If and f(7) are all equal to zero.

Therefore,

Also

And

Hence, and are the factor of the polynomial f(x).

Question 10:

For what value of a is (x − 5) a factor of x³ − 3x² + ax − 10?

Answer 10:

Let be the given polynomial.

By factor theorem, is the factor of f(x), if f (5) = 0

Therefore,

Hence, a = − 8.

Question 11:

Find the value of a such that (x − 4)  is a factors of 5x³ − 7x² − ax − 28.

Answer 11:

Let be the given polynomial.

By the factor theorem,

(x − 4) is a factor of f(x).

Therefore f(4) = 0

Hence ,

$$\begin{gathered} \Rightarrow 320-112-4a-28=0 \\ \Rightarrow 180-4a=0 \\ \Rightarrow a=\frac{180}{4}=45 \end{gathered}$$

Hence,

Question 12:

Find the value of a, if x + 2 is a factor of 4x⁴ + 2x³ − 3x² + 8x + 5a.

Answer 12:

Let 4x⁴ + 2x³ − 3x² + 8x + 5a be the polynomial.

By the factor theorem,

is a factor of f(x) if f(−2) = 0.

Therefore,

Hence,

Question 13:

Find the value k if x − 3 is a factor of k²x³ − kx² + 3kx − k.

Answer 13:

Let be the given polynomial.

By the factor theorem,

(x − 3) is a factor of f(x) if f (3) = 0

Therefore,

$$\begin{gathered} \Rightarrow 27k^2-9k+9k-k=0 \\ \Rightarrow 27k^2-k=0 \\ \Rightarrow k\left(27k-1\right)=0 \\ \Rightarrow k=0 or k=\frac{1}{27} \end{gathered}$$

Hence, the value of k is 0 or .

Question 14:

Find the values of a and b, if x² − 4 is a factor of ax⁴ + 2x³ − 3x² + bx − 4

Answer 14:

Let and be the given polynomial.

We have,

are the factors of g(x).

By factor theorem, if and both are the factor of f(x)

Then f(2) and f(−2) are equal to zero.

Therefore,

and

Adding these two equations, we get

Putting the value of a in equation (i), we get

Hence, the value of a and b are 1, − 8 respectively.

Question 15:

Find α and β, if x + 1 and x + 2 are factors of x³ + 3x² − 2αx + β.

Answer 15:

Let be the given polynomial.

By the factor theorem, and are the factor of the polynomial f(x) if and both are equal to zero.

Therefore,

$$\begin{gathered} f(-1)=(-1{)}^3+3(-1{)}^2-2\alpha \left(-1\right)+\beta =0 \\ \Rightarrow f(-1)=-1+3+2\alpha +\beta =0 \\ \Rightarrow 2\alpha +\beta =-2 ...\left(\mathrm{i}\right) \end{gathered}$$

and

Subtracting (i) from (ii)

We get,

                                 $\alpha =-1$

Putting the value of $\alpha$ in equation (i), we get

Hence, the value of α and β are −1, 0 respectively.

Question 16:

If x − 2 is a factor of each of the following two polynomials, find the values of a in each case

(i) x³ − 2ax² + ax − 1

(ii) x⁵ − 3x⁴ − ax³ + 3ax² + 2ax + 4

Answer 16:

(i) Let be the given polynomial.

By factor theorem, if (x − 2) is a factor of f(x), then f (2) = 0

Therefore,

Thus the value of a is 7/6.

(ii) Let f(x) = x⁵ − 3x⁴ − ax³ + 3ax² + 2ax + 4 be the given polynomial.

By the factor theorem, (x − 2) is a factor of f(x), if f (2) = 0

Therefore,

Thus, the value of a is .

Page-6.25

Question 17:

In each of the following two polynomials, find the value of a, if x − a is  factor:

(i) x⁶ − ax⁵ + x⁴ − ax³ + 3x − a + 2

(ii x⁵ − a²x³ + 2x + a + 1)

Answer 17:

(i) Let be the given polynomial.

By factor theorem, (x − a) is a factor of the polynomial if f(a) = 0

Therefore,

Thus, the value of a is − 1.

(ii) Let be the given polynomial.

By factor theorem, (x − a) is a factor of f(x), if f(a) = 0.

Therefore,

Thus, the value of a is − 1/3.

Question 18:

In each of the following two polynomials, find the value of a, if x + a is a factor.

(i) x³ + ax² − 2x +a + 4

(ii) x⁴ − a²x² + 3x −a

Answer 18:

(i) Let be the given polynomial.

By the factor theorem, (+ a) is the factor of f(x), if f(− a) = 0, i.e.,

Thus, the value of a is − 4/3.

(ii) Let be the polynomial. By factor theorem, (x + a) is a factor of the f(x), if f(− a) = 0, i.e.,

Thus, the value of a is 0.

Question 19:

Find the values of p and q so that x⁴ + px³ + 2x³ − 3x + q is divisible by (x² − 1).

Answer 19:

Let and be the given polynomials.

We have,

Here, are the factor of g(x).

If f(x) is divisible by and , then and are factor of f(x).

Therefore, f(1) and f(−1) both must be equal to zero.

Therefore,

and

Adding both the equations, we get,

Putting this value in (i)

Hence, the value of p and q are 3, −3 respectively.

Question 20:

Find the values of a and b so that (x + 1) and (x − 1) are factors of x⁴ + ax³ − 3x² + 2x + b.

Answer 20:

Let be the given polynomial.

By factor theorem, and are the factors of f(x) if f(−1) and f(1) both are equal to zero.

Therefore,

and

Adding equation (i) and (ii), we get

Putting this value in equation (i), we get,

Hence, the value of a and b are – 2 and 2 respectively.

Question 21:

If x³ + ax² − bx+ 10 is divisible by x² − 3x + 2, find the values of a and b.

Answer 21:

Let and be the given polynomials.

We have,

Here, and are the factors of g(x),

Now,

By factor theorem,

and

Subtracting (ii) by (i), we get,

$$\begin{gathered} \left(2a-b\right)-\left(a-b\right)=-9-\left(-11\right) \\ a=2 \end{gathered}$$
Putting this value in equation (ii), we get,

Hence, the value of a and b are 2 and 13 respectively.

Question 22:

If both x + 1 and x − 1 are factors of ax³ + x² − 2x + b, find the values of a and b.

Answer 22:

Let be the given polynomial.

By factor theorem, if and both are factors of the polynomial f (x). if f(−1) and f(1) both are equal to zero.

Therefore,

And

Adding (i) and (ii), we get

And putting this value in equation (ii), we get,

a = 2

Hence, the value of a and b are 2 and −1 respectively.

Question 23:

What must be added to x³ − 3x² − 12x + 19 so that the result is exactly divisibly by x² + x - 6 ?

Answer 23:

Let and be the given polynomial.

When p(x) is divided by q(x), the reminder is a linear polynomial in x.

So, let r(x) = ax + b is added to p(x), so that p(x) + r(x) is divisible by q(x).

Let

Then,

We have,

Clearly, q(x) is divisible by and i.e., and are the factors of q(x).

Therefore, f (x) is divisible by q(x), if and are factors of f(x), i.e.,

and f(2) = 0

Now, f(-3) = 0

$\Rightarrow$ f(-3) = (-3)³ -3(-3)² + (a-12)(-3)+19+b = 0

$\Rightarrow$ -27 - 27 - 3a + 36 + 19 + b = 0

$\Rightarrow$ -27 - 27 - 3a + 36 + 19 + b = 0

$\Rightarrow$ -54 - 3a + b + 55 = 0

$\Rightarrow$ -3a + b + 1 = 0    ---- (i)
 

And

Subtracting (i) from (ii), we get,

Putting this value in equation (ii), we get,

Hence, p(x) is divisible by q(x) if added to it.

Question 24:

What must be subtracted from x³ − 6x² − 15x + 80 so that the result is exactly divisible by x² + x − 12?

Answer 24:

By divisible algorithm, when is divided by the reminder is a linear polynomial

Let be subtracted from p(x) so that the result is divisible by q(x).

Let

We have,

Clearly, and are factors of q(x), therefore, f(x) will be divisible by q(x) if and are factors of f(x), i.e. f (−4) and f (3) are equal to zero.

Therefore,

and

Adding (i) and (ii), we get,

Putting this value in equation (i), we get,

Hence, will be divisible by if 4 x − 4 is subtracted from it

Question 25:

What must be added to 3x³ + x² − 22x + 9 so that the result is exactly divisible by 3x² + 7x − 6?

Answer 25:

By division algorithm, when is divided by the reminder is a linear polynomial. So, let r(x) = ax + b be added to p(x) so that the result is divisible by q(x)

Let

We have,

$$\begin{gathered} q\left(x\right) = 3x^2 + 7x - 6 \\ = 3x^2 + 9x - 2x - 6 \\ = 3x\left(x + 3\right) - 2\left(x + 3\right) \\ = \left(3x - 2\right) \left(x + 3\right) \end{gathered}$$

Clearly, $\left(3x - 2\right)$ and $\left(x + 3\right)$ are factors of q(x).

Therefore, f(x) will be divisible by q(x) if and are factors of f(x), i.e.,

and f(−3) are equal to zero.

Now,

$$\begin{gathered} f\left(\frac{2}{3}\right) = 0 \\ \Rightarrow 3{\left(\frac{2}{3}\right)}^3 + {\left(\frac{2}{3}\right)}^2 + \left(a - 22\right)\left(\frac{2}{3}\right) + 9 + b = 0 \\ \Rightarrow 3\times \frac{8}{27} + \frac{4}{9} + \frac{2a}{3} - \frac{44}{3} + 9 + b = 0 \\ \Rightarrow \frac{8}{9} + \frac{4}{9} - \frac{44}{3}+ 9 + \frac{2a}{3} + b = 0 \\ \Rightarrow \frac{8 + 4 - 132 + 81 }{9} + \frac{2a}{3} + b = 0 \\ \Rightarrow -\frac{39}{9} + \frac{2a}{3} + b = 0 \\ \Rightarrow \frac{2a}{3} + b = \frac{13}{3} \\ \Rightarrow 2a + 3b = 13 ........\left(\mathrm{i}\right) \end{gathered}$$

And

$$\begin{gathered} f\left(-3\right) = 0 \\ \Rightarrow 3{\left(-3\right)}^3 + {\left(-3\right)}^2 + \left(a - 22\right)\left(-3\right) + 9 + b = 0 \\ \Rightarrow -81 + 9 - 3a + 66 + 9 + b = 0 \\ \Rightarrow -3a + b = -3 \\ \Rightarrow b = -3 + 3a .........\left(\mathrm{ii}\right) \end{gathered}$$

Substituting the value of b from (ii) in (i), we get,
$$\begin{gathered} 2a + 3\left(3a - 3\right) = 13 \\ \Rightarrow 2a + 9a - 9 = 13 \\ \Rightarrow 11a = 13 + 9 \\ \Rightarrow 11a = 22 \\ \Rightarrow a = 2 \end{gathered}$$
Now, from (ii), we get
$b = - 3 + 3\left(2\right) = -3 + 6 = 3$

So, we have a = 2  and  b = 3

Hence, p(x) is divisible by q(x), if is added to it.