Question

If 2x²+ax²+bx-2 has a factor (x+2) and leaves a remainder 7 when divided by 2x-3, find the values of a and b. With these values of a and b, factorise the given polynomial compleatly.

Answer 1

The values of a and b are  3 , -3 respectively

Therefore the factorised given polynomial can be written as

$P(X)=2x^3+3x^2-3x-2=(x+2)(2x+1)(x-1)$

Step-by-step explanation:

Given that $2x^3+ax^2+bx-2$ has a factor (x+2) and leaves a remainder 7 when divided by 2x-3

To find the values of a and b and also factorise the given polynomial completely :

Let P(X) be the given polynomial

$P(X)=2x^3+ax^2+bx-2$

Since the given polynomial has a factor x+2

x+2=0

x=-2

Therefore x=-2 is the zero for P(X)=0

That is put x=-2 in P(X) we get

$P(-2)=2(-2)^3+a(-2)^2+b(-2)-2=0$

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

$-16+4a-2b-2=0$

$4a-2b-18=0$

$4a-2b=18\hfill (1)$

From the given the given polynomial leaves a remainder 7 when divided by 2x-3

2x-3=0

2x=3

$x=\frac{3}{2}$ leaves remainder 7

Hence put $x=\frac{3}{2}$ in P(X)

$P(\frac{3}{2})=2(\frac{3}{2})^3+a(\frac{3}{2})^2+b(\frac{3}{2})-2=7$

$2(\frac{3^3}{2^3})+a(\frac{3^2}{2^2})+\frac{3b}{2}=7+2$

$\frac{27}{4}+a(\frac{9}{4})+\frac{3b}{2}=9$

$\frac{27+9a+3b(2)}{4}=9$

$27+9a+6b=9\times 4$

$9a+6b=36-27$

$9a+6b=9\hfill (2)$

Now solving the equations (1) and (2) we get

Multiply the equation (1) into 3 we get

$12a-6b=54\hfill (3)$

Now adding the equations (2) and (3)

$9a+6b=9$

$12a-6b=54$

___________________

$21a=63$

$a=\frac{63}{21}$

$=3$

Therefore the value of a is 3

• Now substitute the value of a=3 in the equation (1) we get

$4(3)-2b=18$

$-2b=18-12$

$b=-\frac{6}{2}$

$=-3$

Therefore the value of b is -3

Now substitute the values of a and b in the given polynomial P(X)

$2x^3+3x^2+(-3)x-2$

$2x^3+3x^2-3x-2=0$

By using the synthetic division we can factorise it

-2 |  2      3      -3      -2

      0      -4      2       2

    ________________

     2       -1      -1        0

Therefore x+2 is a factor

Therefore x=-2 is a zero of P(X)

Now we have the quadratic equation $2x^2-x-1=0$

$2x^2-2x+x-1=0$

$2x(x-1)+1(x-1)=0$

$(2x+1)(x-1)=0$

2x+1=0 or  x-1=0

$x=-\frac{1}{2}$ and x=1 are the zeros of P(X)

Therefore the factorised given polynomial can be written as

$P(X)=2x^3+3x^2-3x-2=(x+2)(2x+1)(x-1)$

The factors of the given polynomial are (x+2) , (2x+1) and (x-1)