Question

c) If 2x' + ax + bx + (6) has a factor (2x + 1) and leaves a remainder 12 when divided by (x + 2), calculate the
values of a and b. Hence, factorise the given expression completely.​

Answer 1

Answer:

a=3,b=-11

and factors are:

(2x+1)(x+3)(x-2)

Step-by-step explanation:

If 2x³ + ax² + bx -6 has a factor (2x + 1) and leaves a remainder 12 when divided by (x + 2), calculate the  values of a and b. Hence, factorize the given expression completely.

let p(x)=2x³ + ax² + bx -  6

2x+1 is a factor of p(x)

2x+1=0, x=-1/2

so p(-1/2)=0

2(-1/2)³+a(-1/2)²+b(-1/2)-6=0

-1/4+a/4-b/2-6=0

-1+a-2b-24=0

a-2b=25-----------------(1)

Now when P(x) is divided by x+2 then remainder is 12

Thus p(-2)=12

2(-2)³+4a-2b-6=12

-16+4a-2b-6=12

4a-2b=34-------------(2)

Subtracting(1) from(2)

3a=9,a=3

from(1)

3-2b=25

-2b=22,b=-11

Thus p(x)=2x³+3x²-11x-6

=2x³+x²+2x²+x-12x-6

=x²(2x+1)+x(2x+1)-6(2x+1)

=(2x+1)(x²+x-6)

=(2x+1)( x²+3x-2x-6)

=(2x+1)(x(x+3)-2(x+3))

=(2x+1)(x+3)(x-2)