About Factoring by Grouping:
In some cases, we will need to factor a four-term polynomial by grouping. When we factor by grouping, we are arranging our polynomial into two groups with two terms in each group. From each group, we will pull out the GCF or the -GCF and look for a common binomial factor. Sometimes, we need to rearrange our terms or try a different grouping in order to obtain a common binomial factor.
Test Objectives
- Demonstrate an understanding of how to find the GCF
- Demonstrate the ability to factor out the GCF
- Demonstrate the ability to factor a four-term polynomial using Grouping
#1:
Instructions: Factor each.
$$a)\hspace{.2em}x^3 + 5x^2 + 5x + 25$$
$$b)\hspace{.2em}9x^3 + 3x^2 + 15x + 5$$
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#2:
Instructions: Factor each.
$$a)\hspace{.2em}4x^3 - 10x^2 + 6x - 15$$
$$b)\hspace{.2em}5xy - 3x + 25y - 15$$
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#3:
Instructions: Factor each.
$$a)\hspace{.2em}60xy - 96x - 90y + 144$$
$$b)\hspace{.2em}48xy + 42x^2 + 40y + 35x$$
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#4:
Instructions: Factor each.
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$$a)\hspace{.2em}60xy - 112a^2 + 140xa - 48ay$$
$$b)\hspace{.2em}64x^2y + 80x^2m + 128x^3 + 40xmy$$
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#5:
Instructions: Factor each.
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$$a)\hspace{.2em}196xy - 48n^2 - 56xn + 168ny$$
$$b)\hspace{.2em}14y^2xc + 24y^3f - 56y^2xf - 6y^3c$$
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Written Solutions:
#1:
Solutions:
$$a)\hspace{.2em}(x^2 + 5)(x + 5)$$
$$b)\hspace{.2em}(3x^2 + 5)(3x + 1)$$
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#2:
Solutions:
$$a)\hspace{.2em}(2x^2 + 3)(2x - 5)$$
$$b)\hspace{.2em}(x + 5)(5y - 3)$$
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#3:
Solutions:
$$a)\hspace{.2em}6(2x - 3)(5y - 8)$$
$$b)\hspace{.2em}(6x + 5)(8y + 7x)$$
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#4:
Solutions:
$$a)\hspace{.2em}4(5x - 4a)(3y + 7a)$$
$$b)\hspace{.2em}8x(8x + 5m)(2x + y)$$
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#5:
Solutions:
$$a)\hspace{.2em}4(7x + 6n)(7y - 2n)$$
$$b)\hspace{.2em}2y^2(7x - 3y)(c - 4f)$$
