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Oh, what a happy little math problem we have here! To factor w^2 + 8w + 12, we're looking for two numbers that multiply to 12 and add up to 8. Those numbers are 2 and 6. So, we can rewrite the equation as (w + 2)(w + 6). Just like that, we've created a beautiful factorization!

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BobBot

4mo ago

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w2 + 8w + 12 = (w + 6) (w + 2)

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Wiki User

14y ago
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(w+ 6)(w + 2)

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Wiki User

9y ago
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Q: How do you factor w2 plus 8w plus 12?
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How can you factor this trinomial w2 plus 5w - 14?

(w + 7)(w - 2)


Reduce the fraction w2 plus 5w plus 6 over w2-w-12?

Ah, math time, my favorite! To reduce the fraction w^2 + 5w + 6 over w^2 - w - 12, first factor both the numerator and the denominator. The numerator factors into (w + 2)(w + 3), and the denominator factors into (w + 3)(w - 4). Cancel out the common factor of (w + 3) in both the numerator and the denominator, leaving you with (w + 2) over (w - 4). Voilà!


What is ww plus 97 equals 133?

It is either a linear equation in a variable ww or a quadratic equation in a variable w, (with ww denoting w2).


A triangle is 3cm wider than it tall the area is 27cm2find the width?

Area of triangle = 0.5 * width * height So 27 = 0.5 * w * (w-3) or 54 = w2 - 3w ie w2 - 3w - 54 = 0 w2 - 9w + 6w - 54 = 0 w(w - 9) + 6(w - 9) = 0 ie (w - 9)(w + 6) = 0 So w = 9 or w = -6 But since w is a width, it must be positive and so w = 9 cm.


What are the dimensions of rectangle if the diagonal is 13cm?

The length of the diagonal is not enough to determine the dimensions of the rectangle. Take any length W such that 0<W<sqrt(132/2) that is, 0<W<9.192 cm. And let L = sqrt(132 - W2) cm. Then 9.192<L<13 so that any combination of W and L is unique. There are infinitely many possible values for W and so infinitely many different combinations for W and L. That is, infinitely many rectangles whose width is W cm and length is L cm. And, since, L2 = 132 - W2, L2 + W2 = 132 so the diagonal is 13 cm.