Find a and b given $(3x + 2)$ and $(x - 2)$ are factors of $6x^3 + ax^2 - 4ax + b$.

Question

$(3x + 2)$ and $(x - 2)$ are factors of $6x^3 + ax^2 - 4ax + b$. Find a and b.

Answer 1

Multiply $(3x+2)(x-2)(2x+r)$ out. Then equate and solve for a and b.

Answer 2

As J. W. Tanner wrote, putting $x = 2$ gives $6x^3 + ax^2 - 4ax + b =6 8 + 4a - 8a + b =b-4a+48 = 0 $ so $4a-b = 48 $.

Putting $x = -\frac23$ gives $6x^3 + ax^2 - 4ax + b =-6 \frac{8}{27} + \frac49 a +\frac83 a + b =- \frac{16}{9} + \frac{28}{9} a + b $ so $28a+9b = 16 $.

$28a-7b = 48\ 7 = 336 $ so $16b = -320, b = -20 $, $a = (48+b)/4 =7 $.

And, according to Wolfy, $$\frac {6x^3 + 7x^2 - 28x -20}{(3x+2)(x-2)} =2x+5 $$.

Answer 3

Note

$$f(x)=6x^3 + ax^2 - 4ax + b=(3x + 2)(x - 2)(2x-r)$$

Set

$$f(0)= b=4r,\>\>\>\>\> f(4)=384+b=28(8-r)$$

to get $b=-20$ and $r=-5$. As a result

$$f(1)= -3a-14=-35\implies a=7$$