Question 907430
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You do not have anything that can be "solved".  In order to "solve" something you need some sort of relationship operator, such as an equals sign or an inequality sign.  I go to the trouble of saying this because precision of language is essential in mathematics.


Fortunately, you don't need to solve anything to find the information that is requested.  What you need to do is to expand the binomial expressions and collect like terms.  And even more fortunately, you don't even need to completely perform this operation.


Your factored form polynomial:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  f(x)\ =\ -10(x\ +\ 5)^2(x\ +\ 1)(x\ -\ 4)^3]


If you expand the *[tex \Large (x\ +\ 5)^2] binomial, you will get *[tex \Large x^2] plus some lower order terms forming a quadratic trinomial.  If you expand *[tex \Large (x\ -\ 4)^3], you get *[tex \Large x^3] plus some lower order terms forming a four-term cubic polynomial.  Then, multiplying a trinomial with an *[tex \Large x^2] high order term times a degree 1 binomial, *[tex \Large (x\ +\ 1)], times a four term polynomial with a high order term of *[tex \Large x^3] you will get a polynomial with a high order term of *[tex \Large x^6].  Hence, the polynomial is degree 6.


Then, you multiply through by *[tex \Large -10] giving you a high-order term of *[tex \Large -10x^6], and by inspection, the lead coefficient is *[tex \Large -10].


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
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