homework help 22354

Use the new AC Method.

Case 1. Factoring trinomial type ##f(x) = x^2 + bx + c##.

The factored trinomial will have the form: ##f(x) = (x + p)(x + q)##.

The new AC Method finds ##2## numbers ##p and q## that satisfy these 3 conditions:

1. The product ##p*q = a*c##. (When ##a = 1##, this product is ##c##)
2. The sum ##(p + q) = b##
3. Application of the rule of Signs for real roots.

Reminder of the Rule of Signs.

• When ##a and c## have different signs, ##p and q## have opposite signs.
• When ##a and c## have the same sign, ##p and q## have the same sign.

New AC Method.

To find ##p and q##, compose factor pairs of ##c##, and in the same time, apply the Rule of Signs . The pair whose sum equals to ##(-b)##, or ##(b)##, gives ##p and q##.

Example 1. Factor ##f(x) = x^2 + 31x + 108. ##

Solution. ##p and q## have the same sign. Compose factor pairs of ##c = 108##. Proceed: ##…(2, 54), (3, 36), (4, 27)##. The last sum is ##4 + 27 = 31 = b##. Then, ##p = 4 and q = 27##.
Factoring form: ##f(x) = (x + 4)(x + 27)##
.CASE 2 . Factor trinomial standard type ##f(x) = ax^2 + bx + c## (1)

Bring back to Case 1.

Convert ##f(x)## to ##f'(x) = x^2 + bx + a*c = (x + p’)(x + q’)##. Find ##p’ and q’## by the method mentioned in Case 1 .
Then divide ##p’ and q’## by ##(a)## to get ##p and q## for trinomial (1).

Example 2 . Factor ##f(x) = 8x^2 + 22x – 13 = 8(x + p)(x + q)## (1).

Converted trinomial:

##f'(x) = x^2 + 22x – 104 = (x + p’)(x + q’)## (2).

##p’ and q’## have opposite signs. Compose factor pairs of ##(ac = -104) –> … (-2, 52), (-4, 26)##. This last sum is ##(26 – 4 = 22 = b)##. Then, ##p’ = -4 and q’ = 26##.

Back to the original trinomial (1):

##p = (p’)/a = -4/8 = -1/2 and q = (q’)/a = 26/8 = 13/4##.

Factoring form

##f(x) = 8(x – 1/2)(x + 13/4) = (2x – 1)(4x + 13).##

This new AC Method avoids the lengthy .