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Factor a^2-c^2+b^2+2ab?

Factor completely. Your answer must be expressed as a product of two polynomials.

I would greatly appreciate if you show how you get your answer.

I don't know if this helps, but I've noticed that this equation can be made into the law of cosines...

2 Answers

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  • 1 decade ago
    Favorite Answer

    Move the similar terms together and you should see a pattern:

    a^2 + 2ab + b^2 - c^2

    The a and b terms form a perfect square, with the formula (a+b)^2 = a^2 + 2ab + b^2.

    (a+b)^2 - c^2

    This leaves you with a difference of squares, whose formula is (a+b)(a-b) = a^2 - b^2.

    (a+b+c)(a+b-c)

  • ?
    Lv 7
    1 decade ago

    cute

    your teacher hates you

    LOL

    you are going to end up with a 'difference of squares'

    which become 'two polynomial'

    never mind the polynomial has more than what you think it should

    so,

    a^2-c^2+b^2+2ab

    move -c^2 to the end

    a^2+b^2+2ab-c^2

    rewrite as

    a^2+2ab+b^2-c^2

    factor a^2+2ab+b^2

    (a+b)^2

    gives you

    (a+b)^2-c^2

    difference of squares

    ((a+b)+c)*((a+b)-c)

    (a+b+c)*(a+b-c)

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