Question

If polynomial p (x) = x^3 + ax^2 + bx - 84 is exactly divisible by x^2 + x-12, find the values of a and b.​

Answer 1

Step-by-step explanation:

x²+x-12 is a factor of this polynomial

x²+x-12=0

2x²-12=0

2x²=12

x²=12/2

x²=6

x=√6

PUT THE VALUES AS GIVEN

p(x)=x³+ax²+bx-84=0

p(√6)=(√6)³+a(√6)²+b(√6)-84=0

p(√6)= 6√6+6a+b√6-84=0

p(√6)= √6+6a+b√6=84/6

p(√6)= √6+6a+b√6=24

p(√6)=√6+a+b√6=24/6

p(√6)= √6+a+b√6=4

Answer 2

Step-by-step explanation:

Given:If

$p(x) = {x}^{3} + a {x}^{2} + bx - 84$

is completely divisible by

${x}^{2} + x - 12$

To find: Find the value of a and b.

Solution:

Divide p(x) by $x^2+x-12$

${x}^{2} + x - 12) {x}^{3} + a {x}^{2} + bx - 84(x + (a - 1) \\ {x}^{3} + {x}^{2} - 12x\:\: \: \:\:\:\:\:\:\:\:\\ - \:\:\: \: \:\:\: \: \: \: - \:\: \: \: \: \: \: + \: \:\:\: \\ - - - - - - - - - \\ (a - 1) {x}^{2} + (b + 12)x - 84\:\:\:\: \\ (a - 1) {x}^{2} + (a - 1)x - 12(a - 1) \\ - \: \: \: \: \: \: \: \: \: \: - \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: + \: \: \: \: \: \: \: \: \: \: \: \\ - - - - - - - - - - - - - - - \\ (b + 12 - a + 1)x - 84 + 12a - 12$

if p(x) is divisible by $x^2+x-12$ then remainder will be zero

$(b - a + 13)x -96 + 12a = 0$

remainder will be zero only if

$b - a + 13 = 0 \\ \\ b - a = - 13...eq1$

or if

$-96 + 12a = 0 \\ \\ 12a = 96 \\ \\ a = \frac{96}{12} \\ \\ a = 8$

put the value of a in eq1

$b - 8= - 13 \\ \\ b = - 13+8 \\ \\ b = - 5$

Final answer:

$\bold{a = 8} \\ \\ \bold{b = -5 }$

Hope it helps you.

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