McGraw Hill Integrated II, 2012
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McGraw Hill Integrated II, 2012 View details
Study Guide and Review
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Exercise 13 Page 217

Is there a GCF between all of the terms in the given expression? If so, you should factor that out first.

(x-5)(3x-1)

Practice makes perfect

Factor the Quadratic Trinomial

Here we have a quadratic trinomial of the form ax^2+bx+c, where |a| ≠ 1 and there are no common factors. To factor this expression, we will rewrite the middle term, bx, as two terms. The coefficients of these two terms will be factors of ac whose sum must be b. 3x^2-16x+5 ⇔ 3x^2+(- 16)x+5 We have that a= 3, b=- 16, and c=5. 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=5, the value of a c is 3* 5=15.
  1. Find factors of a c. Since a c=15, which is positive, we need factors of a c to have the same sign — both positive or both negative — in order for the product to be positive. Since b=- 16, which is negative, those factors will need to be negative so that their sum is negative.

c|c|c|c 1^(st)Factor &2^(nd)Factor &Sum &Result -3 &-5 &-3 + (-5) & -8 -1 & - 15 & -1 + ( - 15) &- 16

  1. Rewrite bx as two terms. Now that we know which factors are the ones to be used, we can rewrite bx as two terms. 3x^2+(- 16)x+5 ⇔ 3x^2 - 1x - 15x+5
Finally, we will factor the last expression obtained.
3x^2-1x-15x+5
(x(3x-1)-15x+5)
( x(3x-1)-5(3x-1) )
(x-5)(3x-1)

Checking Our Answer

Check your answer âś“
We can expand our answer and compare it with the given expression.
(x-5)(3x-1)
3x (x-5) -1 (x-5)
3 x^2 - 15 x -1 (x-5)
3 x^2 - 15 x - x + 5
3 x^2 - 16 x -5
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!