Question

If polynomial p(x) = x3+ax2+bx-84 is exactly divisible by x2+x-12, find the values of a and b​

Answer 1

Answer:

A is 4th polynomial and

B is 2nd polynomial

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 $x^2+x-12$

${x}^{2} + x - 12) {x}^{3} + a {x}^{2} + bx - 84(x + (a - 1) \\ {x}^{3} + {x}^{2} - 12x\:\: \:\:\:\\( -) \:\:\:( -) \:\: \: \:(+) \: \:\:\:\quad \quad\\ - - - - - - - - - \\ (a - 1) {x}^{2} + (b + 12)x - 84\:\:\:\:\:\:\:\:\: \\ (a - 1) {x}^{2} + (a - 1)x - 12(a - 1) \\ (-) \: \: \: \: \: \: (-) \: \: \:\: \: \: \: \: \: \: \: \: \: (+) \: \: \: \: \: \: \: \quad\\ - - - - - - - - - - - - - - - \\ \bold{(b - a + 13)x + 12a-96} \\- - - - - - - - - - - - - - -$

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

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

remainder will be zero only if either the coefficient of x will be zero or the constant term will be equal to zero.Thus,

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

or if

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

put the value of a in eq1

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

Final answer:

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

Hope it helps you.

To learn more on brainly:

1) if (X + 2) is a factor of X⁵ - 4a²x + 2 X +2a+3 find a. https://brainly.in/question/12783153

2) If one of the Zeroes of the quadratic polynomials (a-1)x+ ax+1= -3, then find the value of a. https://brainly.in/question/41118278