How do I simplify these fraction?

Question

I thought I was doing them correctly, but I got them wrong. Can some one show me how to do it correctly, thank you.

1. a^2-2ab+b^2 / a+b
2.2x^2+x-6 /2x+4

Answer 1

You have to factor.

1) factored form: (a-b)(a-b) / (a+b)
since nothing cancels, the first one is simplified by multiplying the same terms together to get:
(a-b)^2 / (a+b)

2) factored form: (2x-3)(x+2) / 2(x+2)
the (x+2) on top and bottom cancel, so simplified form is:
(2x-3) / 2

Answer 2

These fractions can only be simplified if the denominator is a factor of the numerator.

1. a + b is not a factor of a^2 - 2ab + b^2, so we cannot simplify. Note that a^2 - 2ab + b^2 = (a - b)^2; a + b is not a factor. One of the answerers below me has factored a^2 - 2ab + b^2 into (a + b)(a - b), but (a + b)(a - b) = a^2 - b^2, not a^2 - 2ab + b^2. Another answerer has mistaken the denominator of a + b for a - b.

2. 2x^2 + x - 6 = (2x + 4)(x - 1.5), so when you divide by 2x + 4 it cancels from the top and bottom, leaving a result of x - 1.5. Also note that (2x - 3)(x + 2) is an incorrect factorization of 2x^2 + x - 6, since (2x - 3)(x + 2) = 2x^2 - x + 6; the first-order term is -x instead of x.

Answer 3

Factor the numerators, then see if you can get some terms to cancel if they show up in both the numerator and denominator. For example, the first one simplifies to (a-b) like so:

(a^2 - 2ab + b^2) / (a+b)
(a+b)(a-b) / (a+b)
a-b

Answer 4

1.(a-b)^2
_________

a+b

2. 2x-6
_____
4

Answer 5

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