Question

if x^3+ax^2-bx+10 is divisible by x^2-3x+2,find the values of a and b

Answer 1

(a,b)≡(2,13)

Consider x2−3x+2.

This can be factorised as (x−1)(x−2).

So, x3+ax2−bx+10 has to be divisible by both (x−1) and (x−2).

By synthetic division, on dividing with (x−1), we get the remainder as a−b+11.

So, we have the first equation: a−b+11=0

And on dividing by (x−2), we get the remainder as 4a−2b+18.

So, the second equation: 2a−b+9=0

So, a+11=2a+9⇒a=2 and b=13.

I hope this helps you. Let us know if there is any confusion!

Answer 2

Step-by-step explanation:

Given: If $p(x) = {x}^{3} + a {x}^{2} - bx +10$is completely divisible by ${x}^{2} -3x+2$.

To find: Find the value of a and b.

Solution:Divide p(x) by $x^2-3x+2$

${x}^{2} -3x+2) {x}^{3} + a {x}^{2} - bx +10(x + (a+3) \\ {x}^{3} -3 {x}^{2} +2x\:\: \:\:\:\\( -) \:\:\:( +) \:\: \: \:(-) \: \:\:\:\quad \quad\\ - - - - - - - - - \\ (a+3) {x}^{2} - (b + 2)x +10\:\:\:\:\:\:\:\:\: \\ (a+3) {x}^{2} -3(a +3)x + 2(a +3) \\ (-) \: \: \: \: \: \: (+) \: \: \:\: \: \: \: \: \: \: \: \: \: (-) \: \: \: \: \: \: \: \quad\\ - - - - - - - - - - - - - - - \\ \bold{(3a-b+7 )x - 2a+4} \\- - - - - - - - - - - - - - -$

if p(x) is divisible by $x^2-3x+2$ then remainder will be zero. Thus,

$(3a - b + 7)x -2a+4 = 0$

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

$3a -b+ 7 = 0 \\ \\ 3a-b= -7...eq1$

or if

$-2a+4 = 0 \\ \\ -2a = -4\\ \\ a = \frac{-4}{-2} \\ \\ a = 2$

put the value of a in eq1

$3(2)-b= - 7 \\ \\ -b = - 7-6 \\ \\ -b = -13\\ \\b=13$

Final answer:

$\bold{a = 2} \\ \\ \bold{b = 13}$

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