Flashcard 12 - Math Flashcard Set for the PERT

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The correct answer is:

A binomial that is of the form \(x^2-y^2\) factors to \((x-y)(x+y)\)
\((x+y)^2 = x^2 + 2xy + y^2\)
\((x-y)^2 = x^2 - 2xy + y^2\)


Explanation:

(a) The difference of the two squares \(x^2\) and \(y^2\): \((x^2 - y^2)\) is factored out as \((x-y)(x+y)\).

(b) The perfect square binomial \((x+y)^2\) is factored out into the trinomial \(x^2 + 2xy + y^2\).

(c) The perfect square binomial \((x-y)^2\) is factored out into the trinomial \(x^2 - 2xy + y^2\).

Once you recognize these special binomials, they become very useful in simplifying equations.

In (b) for instance, the first and last terms of the trinomial are perfect squares of \(x\) and \(y\). The middle term is the product of \(x\) and \(y\) multiplied by \(2\).

Let’s try factoring the square binomial \((3x + 2y)^2\). To get the factored trinomial, square \(3x\) and \(2y\): \(9x^2\) + \(middle\,term\) + \(4y^2\). To get the middle term, multiply: \({3x}\cdot{2y}\cdot{2} = 12xy\).

The factored out trinomial is: \(9x^2 + 12xy + 4y^2\).

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