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Ne4ueva [31]
3 years ago
15

Multiply the expression. see attached picture of problem

Mathematics
1 answer:
zepelin [54]3 years ago
8 0
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.

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