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Pearson Algebra 1 Common Core, 2011

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Pearson Algebra 1 Common Core, 2011 View details

Cumulative Standards Review

Exercise 26 Page 610

To factor a quadratic trinomial of the form ax^2+bx+c, rewrite the middle term, bx, as two terms. The coefficients of these two terms will be factors of ac whose sum must be b.

(x+7)(3x+2)

Practice makes perfect

We want to completely factor the given expression. 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+ 23x+ 14 We have that a= 3, b= 23, and c= 14. There are now three steps we need to follow in order to rewrite the above expression.

Find a c. Since we have that a= 3 and c= 14, the value of a c is 3* 14=42.
Find factors of a c. Since a c=42, 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= 23, which is also positive, those factors will need to be positive so that their sum is positive.

c|c|c|c 1^(st)Factor &2^(nd)Factor &Sum &Result 1 &42 &1 + 42 &43 2 & 21 & 2 + 21 & 23

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+ 23x+14 ⇕ 3x^2 + 2x + 21x+14

Finally, we will factor the last expression obtained.

3x^2+2x+21x+14

Factor out x

x(3x+2)+21x+14

Factor out 7

x(3x+2)+7(3x+2)

Factor out (3x+2)

(x+7)(3x+2)

Checking Our Answer

Check your answer ✓

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

(x+7) (3x+2)

Distribute x+7

3x (x + 7) +2 (x + 7)

Distribute 3 x

3x^2 +21 x +2 (x + 7)

Distribute 2

3x^2 +21 x + 2 x + 14

Add terms

3 x^2 + 23 x + 14

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!