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(x-2)(x-5)
how do you solve that?

2006-12-04 15:46:10 · 5 answers · asked by wat8besha 1 in Science & Mathematics Mathematics

5 answers

You just multiply it through.
1. The first x times the second x
2. The first x times -5
3. The -2 times the second x
4. The -2 times the -5


Make sure you keep track of the positive and negative values, negative times negative equals positive, etc.

Answer is x^2 -7x +10

2006-12-04 15:48:12 · answer #1 · answered by Ciera 3 · 0 0

This is a trick commonly known as FOIL - "first, outside, inside, last".

first: This means you multiply the first terms in each bracket. In this case, it's x times x, or x^2.

outside: Multiply the outside terms; that is, the x in the left bracket and the -5 in the right bracket. It's x times (-5), or -5x

inside: Multiply the inside terms: the second term in the first bracket and the first term in the second bracket. -2 times x = -2x

last: Multiply the second terms of each bracket. -2 times -5 = 10

So we have

x^2 - 5x -2x + 10
x^2 -7x + 10

2006-12-05 00:32:05 · answer #2 · answered by Puggy 7 · 0 0

Foil it out which will be X^2 - 7x + 10

2006-12-04 23:51:12 · answer #3 · answered by supersabin03 1 · 0 0

use distributive property.
x times x, x times -5, -2 times x, -2 times -5


solution : xsquared - 7x + 10

* yeah foil!

2006-12-04 23:50:06 · answer #4 · answered by DisenchantedMe 2 · 0 0

mathematically you cannot solve that.......you can FOIL it............but to solve you need to have an actual equation and not just a polynomial expression

2006-12-04 23:58:45 · answer #5 · answered by anyonebutkc 2 · 0 0

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