Question

Find the values of a and b so that the polynomial x3 + ax2 + bx – 6 is exactly divisible by x2 – 4x + 3.

Answer 1

Answer:

this is the answer of your question.

Answer 2

The given you is to find the values of a and b so that the polynomial x3 + ax2 + bx – 6 is exactly divisible by x2 – 4x + 3.

Let

$p(x) = {x}^{2} - 4x + 3 \\ = {x}^{2} - 3x - x + 3 \\ = x(x - 3) - 1(x - 3) \\ (x - 1)(x - 3)$

$let \: q(x) = {x}^{3} + a {x}^{2} + bx - 6$

p(x) is a factor of q(x), (x-1) and (x-3) are the factors of q(x).

1,3 are the roots of the polynomial q(x).

substitute 1 and 3 in the place of x for q(x).

q(1) =0

$q(1) = 0 \\ {1}^{3} + a( {1}^{2}) + b(1) - 6 = \\ a + b - 5 = 0 \\ a + b = 5$

let the above equation be 1

$q(3) = 0 \\ {3}^{3} + a( {3}^{2} ) +3 b - 6 = 0 \\ 27 + 9a + 3b - 6 = 0 \\ 9a + 3b = 21$

let the above equation be 2.

Now subtract equation 2 from equation 1.

$3a + 3b =1 5 \\ 9a + 3b = - 21 \\- \: \: \: \: \ - \: \: \: \: \: \: + \\ - 6a + 0 = + 36 \\ - 6a = 36 \\ a = - 6$

substitute the value of a in equation 1 we get

$a + b = 5 \\ - 6 + b = 5 \\ b = 5 + 6 = 11$

Therefore , the value of a = -6 and b= 11

# spj2

we can find the similar questions through the link given below.

https://brainly.in/question/17377372?referrer=searchResults