Answer :

Let p (x) = *x*^{3 }- 6*x*^{2 }- 15*x *+ 80 and q (x) = *x*^{2 }+ *x *- 12

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 subtracted to p (x) so that p (x) + r (x) is divisible by q (x).

Let, f (x) = p (x) – r (x)

⇒ f(x) = *x*^{3 }- 6*x*^{2 }- 15*x *+ 80 – (ax + b)

⇒ f(x) = *x*^{3 }- 6*x*^{2 }– (a + 15)*x *+ (80 – b)

We have,

q(x) = *x*^{2 }+ *x *– 12

⇒ q(x) = (x + 4) (x - 3)

Clearly, q (x) is divisible by (x + 4) and (x - 3) i.e. (x + 4) and (x - 3) are factors of q (x)

Therefore, f (x) will be divisible by q (x), if (x + 4) and (x - 3) are factors of f (x).

i.e. f(-4) = 0 and f(3) = 0

f (3) = 0

⇒ (3)^{3} – 6(3)^{2} – 3 (a + 15) + 80 – b = 0

⇒ 27 – 54 – 3a – 45 + 80 – b = 0

⇒ 8 – 3a – b = 0 (i)

f (-4) = 0

⇒ (-4)^{3} – 6 (-4)^{2} – (-4) (a + 15) + 80 – b = 0

⇒ -64 – 96 + 4a + 60 + 80 – b = 0

⇒ 4a – b – 20 = 0 (ii)

Subtract (i) from (ii), we get

⇒ 4a – b – 20 – (8 – 3a – b) = 0

⇒ 4a – b – 20 – 8 + 3a + b = 0

⇒ 7a = 28

⇒ a = 4

Put value of a in (ii), we get

⇒ b = -4

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

r (x) = 4x – 4

Hence, p (x) is divisible by q (x), if r (x) = 4x – 4 is subtracted from it.

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