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Can somebody help me?

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Can somebody help me? [#permalink] New post 16 Feb 2012, 15:46
Can somebody help me with this one

what is (x^2 – 3x)(2x + 5) = 0?

Is it 2x^3+5x^2-6x-15?
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Re: Can somebody help me? [#permalink] New post 16 Feb 2012, 16:28
This would equal 2x^3 + 5x^2 - 6x^2 -15x

Remember to FOIL (first, outer, inner, last) quadratic equations.

This would lead to

(x^2)(2x) + (x^2)(5) + (-3x)(2x) + (-3x)(5) = 2x^3 + 5x^2 - 6x^2 - 15x

First: (x^2)(2x) = 2x^3
Outer: (x^2)(5) = 5x^2
Inner: (-3x)(2x) = -6x^2
Last: (-3x)(5) = -15x
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Re: Can somebody help me? [#permalink] New post 18 Feb 2012, 21:56
Expert's post
You just forgot a power of x. When you multiply (-3x) and (2x), the product is -6x^2. Now we can simplify:

(x^2 – 3x)(2x + 5) = 0

2x^3 + 5x^2 - 6x^2 - 15x = 0

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

The question now is what you want to do with that equation. If you want to solve, none of the above is necessary. Just factor x out of the first term:

(x^2 – 3x)(2x + 5) = 0
x(x-3)(2x+5)=0

x= 0, +3, or -5/2
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Re: Can somebody help me?   [#permalink] 18 Feb 2012, 21:56
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