Question

Multiply the expression. see attached picture of problem

Answer 1

Consider each polynom separately:

1) $3x^2+2x-21$

Since
 $D=2^2-4\cdot 3\cdot (-21)=4+252=256 \\ \sqrt{D}=16 \\ x_{1,2= \dfrac{-2\pm 16}{6} } =-3; \frac{7}{3}$, then $3x^2+2x-21=3(x+3)(x-\frac{7}{3})=(x+3)(3x-7)$.


2)
 $-2x^2-2x+12=2(x+3)(x-2) \\ D=(-2)^2-4\cdot(-2)\cdot 12=4+96=100 \\ \sqrt{D}=10 \\ x_{1,2}=\dfrac{2\pm 10}{-4} =-3;2$.

Then $\dfrac{3x^2+2x-21}{-2x^2-2x+12} = \dfrac{(x+3)(3x-7)}{2(x+3)(x-2) } = \dfrac{3x-7}{2(x-2)}$.

Similarly,3)
 $2x^2+25x+63=2(x+9)(x+\frac{7}{2} ) =(x+9)(2x+7)\\ D=25^2-4\cdot 2\cdot 63=625-504=121 \\ \sqrt{D}=11 \\ x_{1,2}= \dfrac{-25\pm 11}{4} =-9;- \frac{7}{2}$and

4)
 $6x^2+7x-49=6(x- \frac{7}{3})(x+ \frac{7}{2})=(3x-7)(2x+7) \\ D=7^2-4\cdot6\cdot(-49)=1225 \\ \sqrt{D}= 35 \\ x_{1,2}=\dfrac{-7\pm35}{12}= \frac{7}{3} ; -\frac{7}{2}$

Then
 $\dfrac{2x^2+25x+63}{6x^2+7x-49} = \dfrac{(x+9)(2x+7)}{(3x-7)(2x+7) } = \dfrac{x+9}{3x-7}$.
Now multiplication becomes easier:$\dfrac{3x^2+2x-21}{-2x^2-2x+12}\cdot \dfrac{2x^2+25x+63}{6x^2+7x-49} = \dfrac{3x-7}{2(x-2)}\cdot \dfrac{x+9}{3x-7} = \dfrac{x+9}{2(x-2)}$

You obtain answer $\dfrac{x+9}{2x-4}$, where a=1, b=9, c=2, d=-4.