Question

If x2+x-12 divides px =x3+ax2+bx-84 exactly find a nad b

Answer 1

Answer:

Step-by-step explanation:

x2+x-12 =0

x2+ 4x-3x-12=0

x(x+4)-3(x+4)=0

(x-3) (x+4) =0

so x= 3

or. x=-4

p(x)÷g(x)= q(x) + r

p(x)÷g(x)= q(x) +0

x3 + ax2 + bx -84 ÷ x2+x-12

after dividing remainder will be ax2 - x2+bx+12x-84=0

let x=3

27a-9+3b+36-84=0

9a + 3b= 57

3(3a+b)=57

3a+b=19________eq 1

now let x= -4

ax2-x2+bx+12x-84=0

16a-16-4b-48-84=0

16a-4b=148

4(4a-b)=148

4a-b=37_________eq 2

add equation 1 and equation 2

so ,

3a + b +4a -b = 19+37

7a = 56

a= 8

put a=8 in equation 1

3a+b = 19

24 +b=19

or b=-5

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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:

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

Hope it helps you.

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