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Is it possible to factor #y=x^2+15x+36 #? If so, what are the factors?

Answer 1

Factoring #y=x^2+15x+36# gives us #(x+3)(x+12)#

So to factor #y=(x^2+15x+36)#, we need to find values that give us the middle term as we know that x times x gives us #x^2# so that they be in both binomial. So let think of factors that gives us 36 but add up to 15.

6 x 6 = 36 | 6 + 6 = 12 - This set of numbers do not work 18 x 2 = 36 | 18 + 2 = 20 - This set of numbers do not work 12 x 3 = 36 | 12 + 3 = 15 - This set of numbers do work.

So we now can plug in numbers and since #y=x^2+15x+36# has positive values, we do not need to worry about if a number need to have a negative sign or not.

So factoring #y=x^2+15x+36# gives us #(x+3)(x+12)#

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Answer 2

Kennedy Adams

Yes, it is possible to factor the expression y=x^2+15x+36. The factors are (x+3)(x+12).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.