Question

Consider the function P(x)=6x³+ax²+bx-5.
Where a and b are constants.
When P(x) is divided by x+1 there is no remainder
When P(x) is divided by 2x-1 the remainder is -15.

Find the values of a and b.

Answer 1

Answer:

P(x)=6x³+ax²+bx-5

When P(x) is divided by x+1 there is no remainder

=> P( - 1) = 0

$6 {( - 1)}^{3} + a {( - 1)}^{2} + b( - 1) - 5 = 0 \\ - 6 + a - b - 5 = 0 \\ a - b = 11 ...........(1)$

When P(x) is divided by 2x-1 the remainder is -15.

therefore

$P( \frac{1}{2} ) = - 15 \\ 6 {( \frac{1}{2}) }^{3} + a {( \frac{1}{2}) }^{2} + b \frac{1}{2} - 5 = 0 \\ \frac{6}{8} + \frac{a}{4} + \frac{b}{2} - 5 = 0 \\ 6 + 2a + 4b - 40 = 0 \\ 2a + 4b = 34 \\ a + 2b = 17............(2)$

Subtracting (1) from (2)

3b = 6

b = 2

put b = 2 in (1)

a - 2 = 11

a = 13

Answer 2

this

is

correct

...............