Pearson Algebra 1 Common Core, 2011
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Pearson Algebra 1 Common Core, 2011 View details
3. Solving Quadratic Equations
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Exercise 73 Page 566

Start by identifying the values of each term's coefficient.

(3w-2)(w+6)

Practice makes perfect

We want to completely factor the given expression. Here we have a quadratic trinomial of the form aw^2+bw+c, where |a| ≠ 1 and there are no common factors. To factor this expression, we will rewrite the middle term, bw, as two terms. The coefficients of these two terms will be factors of ac whose sum must be b. 3w^2+16w-12 ⇔ 3w^2+16w+(- 12) We have that a= 3, b=16, and c=- 12. There are now three steps we need to follow in order to rewrite the above expression.

  1. Find a c. Since we have that a= 3 and c=- 12, the value of a c is 3* (- 12)=- 36.
  2. Find factors of a c. Since ac=- 36, which is negative, we need factors of a c to have opposite signs — one positive and one negative — in order for the product to be negative. Since b=16, which is positive, the absolute value of the positive factor will need to be greater than the absolute value of the negative factor, so that their sum is positive.

c|c|c|c 1^(st)Factor &2^(nd)Factor &Sum &Result - 1 &36 &-1 + 36 &35 - 2 & 18 & - 2 + 18 &16

  1. Rewrite bw as two terms. Now that we know which factors are the ones to be used, we can rewrite bw as two terms. 3w^2+16w-12 ⇕ 3w^2 - 2w + 18w-12

Finally, we will factor the last expression obtained.

3w^2-2w+18w-12
w(3w-2)+18w-12
w(3w-2)+6(3w-2)
(3w-2)(w+6)

Therefore, we have found that the factored form of the given expression is (3w-2)(w+6).

Checking Our Answer

Check your answer ✓
We can expand our answer and compare it with the given expression.

(3w - 2) (w + 6)
w (3w - 2) +6(3w - 2)
3w^2 - 2w +6(3w - 2)
3w^2 - 2w +18w - 12
3w^2 +16 w -12

We can see above that after expanding and simplifying, the result is the same as the given expression. Therefore, we can be sure our solution is correct!