SOLUTION: Factor completely:
4y^2-12y+9-z^2
So, I tried but I'm stuck. Here's my work so far.
(4y^2-z^2)+(9-12y)
(2y-z)^2 + 3(3-4y)
I'm not sure if it's at all correct.
Algebra.Com
Question 547170: Factor completely:
4y^2-12y+9-z^2
So, I tried but I'm stuck. Here's my work so far.
(4y^2-z^2)+(9-12y)
(2y-z)^2 + 3(3-4y)
I'm not sure if it's at all correct.
Answer by bucky(2189) (Show Source): You can put this solution on YOUR website!
You were on the right track.
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Given to factor:
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Separate the first three terms as follows:
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Note that the terms in the parentheses are an algebraic expression that is a perfect square. It can be factored as shown below:
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Notice now that this new form is the difference between two squares. Recall the rule for the difference of two squares:
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For your problem and
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Substitute these into the rule for the difference of two squares as shown in the steps that follow:
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Remove the interior parentheses on the right side of this equation and you have:
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The right side is the factored form of the problem that you were given. So you can write the answer to this problem in the equation form:
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or maybe you want to rearrange the terms in the parentheses a little so the letters come first as in:
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Whatever your teacher prefers. It's just a matter of preference, but the rearrangement is probably the most commonly used form.
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Hope this helps you understand the problem a little better.
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