Exercise 59 Page 83

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. 2a^2+13a-24 We have that a= 2, b=13, and c=24. There are now three steps we need to follow in order to rewrite the above expression.

Find a c. Since we have that a= 2 and c=- 24, the value of a c is 2* - 24=- 48.
Find factors of a c. Since ac=- 48, 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=13, 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 &48 &- 1 + 48 &47 - 2 &24 &- 2 + 24 &22 - 3 & 16 & - 3 + 16 &13 - 4 &12 &- 4 + 12 &8 - 6 &8 &- 6 + 8 &2

Rewrite bx as two terms. Now that we know which factors are the ones to be used, we can rewrite bx as two terms. 2a^2+13a-24 ⇔ 2a^2 - 3a + 16a-24

Finally, we will factor the last expression obtained.

2a^2-3a+16a-24

FactorOut

Factor out a

a(2a-3)+16a-24

FactorOut

Factor out 8

a(2a-3)+8(2a-3)

FactorOut

Factor out (2a-3)

(a+8)(2a-3)

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

(a+8)(2a-3)

Distr

Distribute (2a-3)

a(2a-3)+8(2a-3)

Distr

Distribute a

2a^2-3a+8(2a-3)

Distr

Distribute a

2a^2-3a+16a-24

SubTerm

Subtract term

2a^2+13a-24

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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