# What Must Be Added to X3 − 3x2 − 12x + 19 So that the Result is Exactly Divisibly by X2 + X - 6 ? - Mathematics

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

#### Solution

Let  p(x) =  x3 − 3x2 − 12x + 19 and q(x) = x^2 + x -6 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  f(x) = p(x) + r(x)

Then,

f(x) = x^3 -3x^2 -12x + 19 +ax + b

 = x^3 -3x^2 + (a-12)x+ (19+b)

We have,

q(x) = x^2 + x-6

 = (x +3)(x - 2)

Clearly, q(x) is divisible by  (x+3)and  (x- 2)i.e.,  (x+3) and (x-2) are the factors of q(x).

Therefore, f (x) is divisible by q(x), if (x + 3) and  (x-2)are factors of f(x), i.e.,

f(-3)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

f(2) = (2)^3 -3(2)^2 + (a-12) + 19 +b = 0

8-12+2a - 24 +19 +b = 0

2a +b = 9            ............. (2)

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

(2a +b)-(-3a + b) = 10

5a = 10

a=2

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

⇒ 2 xx 2 +b = 9

b = 5

Hence, p(x) is divisible by q(x) if  (2x + 5)added to it.

Is there an error in this question or solution?

#### APPEARS IN

RD Sharma Mathematics for Class 9
Chapter 6 Factorisation of Polynomials
Exercise 6.4 | Q 23 | Page 25