When we factor a trinomial into the product of two binomials, we encounter two very different scenarios. The harder scenario occurs when the leading coefficient is not one. When this happens, we generally have two methods available: factoring by grouping and reverse FOIL. In order to use the factoring by grouping method, we must expand our middle term and create a four-term polynomial.

Test Objectives
• Demonstrate the ability to factor out the GCF or -(GCF) from a group of terms
• Demonstrate the ability to factor a trinomial into the product of two binomials
• Demonstrate the ability to factor a trinomial when two variables are involved
Factoring Trinomials AC Method Practice Test:

#1:

Instructions: Factor each.

$$a)\hspace{.2em}7x^2 - 58x + 63$$

$$b)\hspace{.2em}7x^4 + 8x^3 - 12x^2$$

#2:

Instructions: Factor each.

$$a)\hspace{.2em}3x^2 + 19x + 20$$

$$b)\hspace{.2em}5x^3 - 6x^2 - 27x$$

#3:

Instructions: Factor each.

$$a)\hspace{.2em}5x^4 + 28x^3 + 15x^2$$

$$b)\hspace{.2em}{-}60x^2 + 162x + 168$$

#4:

Instructions: Factor each.

$$a)\hspace{.2em}24x^2 - 68x + 48$$

$$b)\hspace{.2em}20x^2 - 75x + 45$$

#5:

Instructions: Factor each.

$$a)\hspace{.2em}50x^2 + 135xy + 90y^2$$

$$b)\hspace{.2em}{-}24x^2 + 138xy - 189y^2$$

Written Solutions:

#1:

Solutions:

$$a)\hspace{.2em}(7x - 9)(x - 7)$$

$$b)\hspace{.2em}x^2(7x-6)(x + 2)$$

#2:

Solutions:

$$a)\hspace{.2em}(3x + 4)(x + 5)$$

$$b)\hspace{.2em}x(5x+9)(x-3)$$

#3:

Solutions:

$$a)\hspace{.2em}x^2(5x+3)(x+5)$$

$$b)\hspace{.2em}{-}6(5x+4)(2x-7)$$

#4:

Solutions:

$$a)\hspace{.2em}4(3x-4)(2x-3)$$

$$b)\hspace{.2em}5(x-3)(4x-3)$$

#5:

Solutions:

$$a)\hspace{.2em}5(2x+3y)(5x+6y)$$

$$b)\hspace{.2em}{-}3(2x-7y)(4x-9y)$$