Consider each polynom separately:
1) [tex]3x^2+2x-21[/tex]
Since
[tex]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} [/tex], then [tex]3x^2+2x-21=3(x+3)(x-\frac{7}{3})=(x+3)(3x-7)[/tex].
2)
[tex]-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[/tex].
Then [tex] \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)} [/tex].
Similarly,3)
[tex]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} [/tex]and
4)
[tex]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} [/tex]
Then
[tex] \dfrac{2x^2+25x+63}{6x^2+7x-49} = \dfrac{(x+9)(2x+7)}{(3x-7)(2x+7) } = \dfrac{x+9}{3x-7} [/tex].
Now multiplication becomes easier:[tex]\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)} [/tex]
You obtain answer [tex] \dfrac{x+9}{2x-4} [/tex], where a=1, b=9, c=2, d=-4.
