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 Sunday 18th of March

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 Depdendent Variable

 Number of equations to solve: 23456789
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 Dependent Variable

 Number of inequalities to solve: 23456789
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# Higher Degrees and Variable Exponents

It is not necessary always to use substitution to factor polynomials with higher degrees or variable exponents. In the next example we use trial and error to factor two polynomials of higher degree and one with variable exponents. Remember that if there is a common factor to all terms, factor it out first.

Example 1

Higher-degree and variable exponent trinomials

Factor each polynomial completely. Variables used as exponents represent positive integers.

a) x8 - 2x4 - 15

b) -18y7 + 21y4 + 15y

c) 2u2m - 5um - 3

Solution

a) To factor by trial and error, notice that x8 = x4 Â· x4. Now 15 is 3 Â· 5 or 1 Â· 15. Using 1 and 15 will not give the required -2 for the coefficient of the middle term. So choose 3 and -5 to get the -2 in the middle term:

x8 - 2x4 - 15 = (x4 - 5)(x4 + 3)

 b) -18y7 + 21y4 + 15y = 3y(6y6 - 7y3 - 5) Factor out the common factor -3y first. = 3y(2y3 + 1)(3y3 - 5) Factor the trinomial by trial and error.

c) Notice that 2u2m = 2um Â· um and 3 = 3 Â· 1. Using trial and error, we get

2u2m - 5um - 3 = (2um + 1)(um - 3).