Question

If (x^3+ax^2+bx+6) has(x-2) as a factor and leaves a remainder 3 when divided by (x-3),find the value of a and



b.

Answer 1

Step-by-step explanation:

${x}^{3} + a {x}^{2} + bx + 6 \: has \: \: (x - 2) \: as \: a \:$

$factor \: and \: leaves \: a \: remainder \: 3 \:$

$when \: divided \: by \: (x - 3)$

Hence substitute x=2 in the polynomial and equate it to 0.

${(2)}^{3} + a {(2)}^{2} + b(2) + 6 = 0 = >$

$8 + 4a + 2b + 6 = 0 = > 4a + 2b +$

$14 = 0 = > 2a + b = - 7...(1)$

Substitute similarly x=3 and equate it to 3.

${(3)}^{3} + a {(3)}^{2} + b(3) + 6 = 3 = >$$27 + 9a + 3b + 6 = 3 = >$

$9a + 3b + 33 - 3 = 0 = > 3(3a + b +$

$10) = 0 = > 3a +b = - 10...(2)$

Subtract (1) from(2) we get

3a+b-(2a+b)= -10+7 = -3 =>a= -3 Hence substitute a=>-3 in the equation (1) =>

3(-3)+b= -10=>b= -10+9= -1. Hope it helps you.