Blog: what is the answer to this problem (34x^15+19x^98)(54x^54+76x^34)

Thursday, March 1, 2012

what is the answer to this problem (34x^15+19x^98)(54x^54+76x^34)

Hello!

It isn't clear what the problem is. I see two variants: extract some more factors (factor completely) and, to the contrast, open the parentheses (express as a sum of monomials). Let's do both tasks.

1. Factoring.

34 and 19 have no common factors (34=2*17), $\displaystyle{{x}}^{{15}}$ and $\displaystyle{{x}}^{{98}}$ have the greatest common factor (GCF) $\displaystyle{{x}}^{{15}}.$ Also, 54 and 76 have GCF 2 (54= $\displaystyle{2}\cdot{{3}}^{{3}},$ 76= $\displaystyle{{2}}^{{2}}\cdot{19}$ ). And GCF of $\displaystyle{{x}}^{{54}}$ and $\displaystyle{{x}}^{{34}}$ is $\displaystyle{{x}}^{{34}}$ (the less power).

And the most factored form is:

$\displaystyle{{x}}^{{15}}{\left({34}+{19}{{x}}^{{83}}\right)}\cdot{2}{{x}}^{{34}}{\left({27}{{x}}^{{20}}+{38}\right)}={2}{{x}}^{{49}}{\left({34}+{19}{{x}}^{{83}}\right)}{\left({27}{{x}}^{{20}}+{38}\right)}.$

2. Making a sum.

To multiply two binomials we have to multiply each term by each and sum the results: (a+b)(u+v)=au+av+bu+bv.

So $\displaystyle{\left({34}{{x}}^{{15}}+{19}{{x}}^{{98}}\right)}{\left({54}{{x}}^{{54}}+{76}{{x}}^{{34}}\right)}=$

$\displaystyle={34}\cdot{54}{{x}}^{{{15}+{54}}}+{34}\cdot{76}{{x}}^{{{15}+{34}}}+{19}\cdot{54}{{x}}^{{{98}+{54}}}\cdot{19}\cdot{76}{{x}}^{{{98}+{34}}}=$