Common Factoring and Grouping is a factoring technique that involves finding the Greatest Common Factor (GCF) of 2 or more monomials (or terms) in order to fully factor the expression.
Process for Factoring a GCF Monomial out of an Expression
- Find the GCF of all the terms in a polynomial. We can do this by factorizing/taking apart each term
- Express each term as a product of the GCF and another factor
- Use distributive property to factor out the GCF
Example
Fully factor the polynomial \(3x + 6y\)
First, we can find the GCF of all the terms in the polynomial by factorizing the terms:
\(3x = 3 \cdot x\)
\(6y = 2 \cdot 3 \cdot y\)
Since we can identify 3 as the GCF, we can now express each term as the product of 3 and another factor:
\(3x = (3)(x)\)
\(6y = (3)(2y)\)
This polynomial can be written as \((3)(x) + (3)(2y)\).
Finally, we can apply distributive property to factor out 3:
\(= 3(x + 2y)\)
Therefore, we can determine that \(3x + 6y\) fully factored is \(3(x + 2y)\).
Fuly factor the following polynomials:
\(17ac - 34ad\)
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First, we can find the GCF of all the terms in the polynomial by factorizing the terms:
\(17ac = 17 \cdot a \cdot c\)
\(34ad = 2 \cdot 17 \cdot a \cdot d\)
Since we can identify 17a as the GCF, we can now express each term as the product of 17a and another factor:
\(17ac = (17a)(c)\)
\(34ad = (17a)(2d)\)
This polynomial can be written as \((17a)(c) - (17a)(2d)\).
Finally, we can apply distributive property to factor out 17a:
\(= 17a(c - 2d)\)
Therefore, we can determine that \(17ac - 34ad\) fully factored is \(17a(c - 2d)\).
\(6n^2p^2 + 12np^2 + 36n^3p^3\)
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First, we can find the GCF of all the terms in the polynomial by factorizing the terms:
\(6n^2p^2 = 2 \cdot 3 \cdot n \cdot n \cdot p \cdot p\)
\(12np^2 = 2 \cdot 2 \cdot 3 \cdot n \cdot p \cdot p\)
\(36n^3p^3 = 2 \cdot 2 \cdot 3 \cdot 3 \cdot n \cdot n \cdot n \cdot p \cdot p \cdot p\)
Since we can identify 6np^2 as the GCF, we can now express each term as the product of 6np^2 and another factor:
\(6n^2p^2 = (6np^2)(n)\)
\(12np^2 = (6np^2)(2)\)
\(36n^3p^3 = (6np^2)(6n^2p)\)
This polynomial can be written as \((6np^2)(n) + (6np^2)(2) + (6np^2)(6n^2p)\).
Finally, we can apply distributive property to factor out \(6np^2\):
\(= 6np^2(n + 2 + 6n^2p)\)
Therefore, we can determine that \(6n^2p^2 + 12np^2 + 36n^3p^3\) fully factored is \(6np^2(n + 2 + 6n^2p)\).
\(3g^2 + 6g + 9\)
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First, we can identify the GCF by factorizing the terms:
\(3g^2 = 3 . g . g\)
\(6g = 2 . 3 . g\)
\(9 = 3 . 3\)
Since we can identify \(3\) as the GCF, we can now express each term as the product of \(3\) and another factor:
\(3g^2 = (3)(g^2)\)
\(6g = (3)(2g)\)
\(9 = (3)(3)\)
This polynomial can be written as \( (3)(g^2) + (3)(2g) + (3)(3) \).
Finally, we can apply distributive property to factor out \(3\):
\(= 3(g^2 + 2g + 3)\)
Therefore, we can determine that \(3g^2 + 6g + 9\) fully factored is \(3(g^2 + 2g + 3)\).
Process for Factoring a GFC Binomial Out Of An Expression
- Identify the GFC
- Divide the GFC out of every term in the polynomial
- Combine both binomials to get the fully factored expression
Example
Fully factor \(2x(x + 7) + 3(x + 7)\)
First, we need to identify the GCF. This is how we get our first binomial:
\(GFC = (x + 7)\)
Next, we need to divide the GCF out of every term in the polynomial. This is how we get the second binomial:
\(= \cfrac{2x(x + 7)}{(x + 7)} + \cfrac{3(x + 7)}{x + 7}\)
\(= 2x + 3\)
We then combine both binomials to get the factored expression:
\(= (2x + 3)(x + 7)\)
Therefore, we can determine that \(2x(x + 7) + 3(x + 7)\) fully factored is \((2x + 3)(x + 7)\).
Fuly factor the following polynomials:
\(4s(r + u) - 3(r + u)\)
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First, we need to identify the GCF. This is how we get our first binomial:
\(GCF = (r + u)\)
Next, we need to divide the GCF out of every term in the polynomial. This is how we get the second binomial:
\(= \cfrac{4s(r + u)}{(r + u)} - \cfrac{3(r + u)}{r + u}\)
\(= 4s + 3\)
We then combine both binomials to get the factored expression:
\(= (4s + 3)(r + u)\)
Therefore, we can determine that \(4s(r + u) - 3(r + u)\) fully factored is \((4s + 3)(r + u)\).
\(2y(x - 3) + 4z(3 - x)\)
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First, we need to identify the GCF This is how we get our first binomial.
We can do this by multiplying the second term by -1:
\(2y(x - 3) - 4z(x - 3)\)
\(GFC = (x - 3)\)
Next, we need to divide the GCF out of every term in the polynomial. This is how we get the second binomial:
\(\cfrac{2y(x - 3)}{(x - 3)} - \cfrac{4z(x - 3)}{x - 3}\)
\(2y - 4z\)
We then combine both binomials to get the factored expression:
\((2y - 4z)(x - 3)\)
Therefore, we can determine that \(2y(x - 3) + 4z(3 - x)\) fully factored is \((2y - 4z)(x - 3)\).
Factoring by Grouping
This is commonly done when there are pairs of terms that can be factored together.
The process for factoring by grouping is:
- Group the first 2 terms toegther and then the last 2 terms together
- Factor out the GFC from each separate binomial
- Factor out the common binomial
Example
Fully factor \(ax + ay + 3x + 3y\)
First, we can group the first and last sets of terms into separate brackets:
\((ax + ay)+(3x + 3y)\)
Next, we can factor out the GFC from each separate binomial:
\(a(x + y) + 3(x + y)\)
Finally, we can factor out the common binomial (in this case \(x + y\)):
\((x + y)(a + 3)\)
Therefore, we can determine that \(ax + ay + 3x + 3y\) fully factored is \((x + y)(a + 3)\).
Fully factor the following expressions by grouping:
\(4x^2 + 6xy + 12y + 8x\)
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First, we can group the first and last pairs of terms into separate brackets:
\((4x^2 + 6xy)+(12y + 8x)\)
Next, we can factor out the GFC from each separate binomial:
\(2x(2x + 3y) + 4(2x + 3y)\)
Finally, we can factor out the common binomial (in this case \(2x + 3y\):
\((2x + 3y)(2x + 4)\)
Therefore, we can determine that \(4x^2 + 6xy + 12y + 8x\) fully factored is \((2x + 3y)(2x + 4)\).
\(y^2 + 3y - ay - 3a\)
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First, we can group the first and last pairs of terms into separate brackets:
\((y^2 + 3y)+(- ay - 3a)\)
Next, we can factor out the GFC from each separate binomial:
\(y(y + 3) - a(y + 3)\)
Finally, we can factor out the common binomial (in this case \(y + 3\):
\((y + 3)(y - a)\)
Therefore, we can determine that \(y^2 + 3y - ay - 3a\) fully factored is \((y + 3)(y - a)\).
The formula for the Surface Area of a rectangular prism is \(SA = 2LW + 2LH + 2WH\).
i. Write this formula in factored form.
ii. If \(L = 10\;[cm]\), \(W = 5\;[cm]\), and \(H = 8\;[cm]\), find the Surface Area using both the original formula and the factored formula. What do you notice? Explain why this is so.
Show Answer
i. All we need to do is factor out the GFC (in this case 2):
\(SA = 2(LW + LH + WH)\)
2. First, we can input the values into the original formula:
\(SA = 2LW + 2LH + 2WH\)
\(SA = 2(10)(5) + 2(10)(8) + 2(5)(8)\)
\(SA = 100 + 160 + 80\)
\(SA = 340\;[cm^2]\)
Next, we can input the values into the factored formula:
\(SA = 2((10)(5) + (10)(8) + (5)(8))\)
\(SA = 2(50 + 80 + 40\)
\(SA = 2(170)\)
\(SA = 340\;[cm^2]\)
Therefore, we can determine that the Surface Area in both the original formula and the factored formula is \(340\;[cm²]\). This is because the formulas are equivalent regardless of their format.