Question

3x^3-2x^2-147x+98=(ax-c)(bx+d)(bx-d) where a,b,c and d positive integers. Work out the value of a,b,c and d

Answer 1

Expand the right side:

$(ax-c)(bx+d)(bx-d)=(ax-c)(b^2x^2-d^2)$

$=ab^2x^3-b^2cx^2-ad^2x+cd^2$

Notice that 98 = 2 * 7 * 7, so we have $c=2$ and $d=7$.

Then

$-ad^2=-147\implies a=\dfrac{147}{49}=3$

and

$ab^2=3\implies b^2=1\implies b=1$