Exercise 17 Page 539

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. 4v^2-16v+7 ⇔ 4v^2+(- 16)v+7 We have that a= 4, b=- 16, and c=7. There are now three steps we need to follow in order to rewrite the above expression.

Find a c. Since we have that a= 4 and c=7, the value of a c is 4* 7=28.
Find factors of a c. Since a c=28, 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 - 1 &- 28 &-1 + (-28) &- 29 - 2 & - 14 & - 2 + ( - 14) &- 16 - 4 &- 7 &-4 + (-7) &- 11

Rewrite bx as two terms. Now that we know which factors are the ones to be used, we can rewrite bx as two terms. 4v^2+(- 16)v+7 ⇕ 4v^2 - 2c - 14v+7

Finally, we will factor the last expression obtained.

4v^2-2v-14v+7

FactorOut

Factor out 2v

2v(2v-1)-14v+7

FactorOut

Factor out - 7

2v(2v-1)-7(2v-1)

FactorOut

Factor out (2v-1)

(2v-7)(2v-1)

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

(2v-7)(2v-1)

Distr

Distribute (2v-1)

2v(2v-1)-7(2v-1)

Distr

Distribute v

4v^2-2v-7(2v-1)

Distr

Distribute -7

4v^2-2v-14v+7

SubTerm

Subtract term

4v^2-16v+7

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!

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