Question

If ax³+bx²+x-6 has x+2 as a factor and leaves remainder 4when divided by x-2 find values of a and b

Answer 1

Step-by-step explanation:

Let p(x) = ax³ + bx² + x - 6

A/C to question,

(x + 2) is the factor of p(x) , and we know this is possible only when p(-2) = 0

So, p(2) = a(-2)³ + b(-2)² - 2 - 6 = 0

⇒-8a + 4b - 8 = 0

⇒ 2a - b + 2 = 0 -------------(1)

again, question said that if we p(x) is divided by ( x -2) then it leaves remainder 4.

so, P(2) = a(2)³ + b(2)² + 2 - 6 = 4

⇒8a + 4b - 4 = 4

2a + b -2 = 0 -------------(2)

solve equations (1) and (2),

4a = 0 ⇒a = 0 and b = 2

Then, equation will be 2x² + x - 6

Answer 2

Answer:

a=0, b=2

Step-by-step explanation:

When divided with x+2 leaves 0 as the remainder(acc. to given information)

sub in the given equation

a(-8)+b(4)+(-2)-6=0

-8a+4b-8=0

-2a+b-2=0

-2a+b=2............(1)

When divided with x-2 leaves the remainder 4

a(8)+b(4)+2-6=4

8a+4b-4=4

8a+4b=8

2a+b=2............(2)

(1)+(2)

2b=4

b=2

2a+2=2

2a=0

a=0

PLEASE MARK THIS ANSWER AS THE BRAINLIEST ANSWER