Question

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

Answer 1

Let p (x) = 3x³ + x² - 22x + 9 and q (x) = 3x² + 7x - 6.

By division algorithm,

When p (x) is divided by q (x), the remainder is a linear expression 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)

= 3x³ + x² – 22x + 9 + (ax + b)

= 3x³ + x² + x (a – 22) + b + 9

We have,

q (x) = 3x² + 7x – 6

q (x) = 3x (x + 3) – 2 (x + 3)

q (x) = (3x – 2) (x + 3)

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

Therefore, f(x) will be divisible by q(x), if (3x – 2) and (x + 3) are factors of f(x). 

i.e. f (2/3) = 0 and f (-3) = 0 [∵ 3x – 2 = 0, x = 2/3 and x + 3 = 0, x = -3] 

f (2/3) = 0

\(\Rightarrow\) \(3(\frac{2}{3})^{3}+(\frac{2}{3})^{2}+\frac{2}{3}(a-2x)+b+9=0\)

\(\Rightarrow\) \(\frac{12}{9}+\frac{2}{3a}-\frac{44}{3}+b+9=0\)

\(\Rightarrow\) \(\frac{12+6a-132+9b+81}{9}=0\)

\(\Rightarrow\) \(6a+9b-39=0\)

⇒ 3 (2a + 3b – 13) = 0

⇒ 2a + 3b – 13 = 0 (i)

Similarly,

f (-3) = 0

⇒ 3 (-3)³ + (-3)² + (-3) (a – 2x) + b + 9 = 0

⇒ -81 + 9 – 3a + 66 + b + 9 = 0

⇒ b – 3a + 3 = 0

⇒ 3 (b – 3a + 3) = 0

⇒ 3b – 9a + 9 = 0 (ii)

Subtract (i) from (ii), we get

3b – 9a + 9 – (2a + 3b – 13) = 0

3b – 9a + 9 – 2a – 3b + 13 = 0

⇒ -11a + 22 = 0

⇒ a = 2

Putting value of a in (i), we get

⇒ b = 3

Putting the values of a and b in r (x) = ax + b, we get

r (x) = 2x + 3

Hence, 

p (x) is divisible by q (x) if r (x) = 2x + 3 is divisible by it.