Question

Factor each trinomial below. Please show your work and check your answer. (1 point each)
1.x2 – 8x + 15
2.a2 – a – 20
3.a2 – 5a – 20
4.a2 + 12ab + 27b2
5.2a2 + 30a + 100

Answer 1

becca, welcome to OpenStudy!!

I'd like to suggest that you post just one question at a time, and also explain as well as y ou are able what you already know about the problem and have tried towards solving it.

But let's get started with your first problem:

Please write it as x^2 -8x +15. that ^ indicates exponentiation.

Ask yourself: What are possible factors of 15? Answer: {-1, -3, -5, 1, 3, 5}.

Ask yourself: which of those 6 possible factors combine to -8?

Answer: -3 -5 = -8.

Therefore, x^2 -8x +15 factors into (x-3)(x-5).

check those factors using the FOIL method for the multiplication of two binomials.

Answer 2

It just may be helpful to you to realize that

x^2 -8x +15 = x^2 - 3x - 5x +15 = x(x-3) - 5(x-3)

and that (x-3) is a factor common to both terms, so that

x(x-3) - 5(x-3) = (x-3)(x-5) after you've factored out x-3.

Answer 3

that's called "factoring by grouping" and is just as valid an approach as is the one I discussed first, above.

Answer 4

x^2 -8x +15 = 0

x^2 -3x -5x +15 = 0

(x^2 - 3x) (-5x + 15) = 0

x(x - 3) -5(x-3) = 0

(x - 3) (x - 5) = 0

Answer 5

gunna do a tricky prob, follow me steps.

a^2 + 12ab + 27b^2 = 0

a^2 + 3ab + 9ab + 27b^2 = 0

(a^2 + 3ab) (9ab + 27b^2) = 0

a(a + 3b) 9b(a+3b) = 0

(a + 3b) (a+9b) = 0

these are your factors, when you expand them, it'll take you back to your original equation. "a^2 + 12ab + 27b^2 = 0"

Answer 6

your last question.

2a^2 + 30a + 100 = 0

2(a^2 + 15a + 50) = 0

2(a^2 + 10a + 5a + 50) = 0

2(a^2 + 10a) (5a + 50) = 0

(2(a(a+10) 5(a + 10) = 0

(a+10) 2(a+5) = 0

when you expand these brackets, it'll take you back to the original equation.

(a+10) 2(a+5) = 0

(a+10) (2a+10) = 0

2a^2 + 10a + 20a + 100 = 0

2a^2 + 30a + 100 = 0