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.
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 \(3\):
Therefore, we can determine that \(3x + 6y\) fully factored is \(\boldsymbol{3(x + 2y)}\).
\(17ac - 34ad\)
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 \(17a\):
Therefore, we can determine that \(17ac - 34ad\) fully factored is \(\boldsymbol{17a(c - 2d)}\).
\(6n^2p^2 + 12np^2 + 36n^3p^3\)
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 \(6np^2\):
Therefore, we can determine that \(6n^2p^2 + 12np^2 + 36n^3p^3\) fully factored is \(\boldsymbol{6np^2(n + 2 + 6n^2p)}\).
\(3g^2 + 6g + 9\)
First, we can identify the GCF by factorizing the terms:
\(3g^2 = 3 \cdot g \cdot g\)
\(6g = 2 \cdot 3 \cdot g\)
\(9 = 3 \cdot 3\)
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 \(3\):
Therefore, we can determine that \(3g^2 + 6g + 9\) fully factored is \(\boldsymbol{3(g^2 + 2g + 3)}\).
Fully factor \(2x(x + 7) + 3(x + 7)\)
First, we need to identify the GCF This is how we get our first binomial:
Next, we need to divide the GCF out of every term in the polynomial. This is how we get the second binomial:
\(= \cfrac{2x(\cancel{x + 7})}{(\cancel{x + 7})} + \cfrac{3(\cancel{x + 7})}{(\cancel{x + 7})}\)
\(= 2x + 3\)
We then combine both binomials to get the factored expression:
Therefore, we can determine that \(2x(x + 7) + 3(x + 7)\) fully factored is \(\boldsymbol{(2x + 3)(x + 7)}\).
\(4s(r + u) - 3(r + u)\)
First, we need to identify the GCF This is how we get our first binomial:
Next, we need to divide the GCF out of every term in the polynomial. This is how we get the second binomial:
\(= \cfrac{4s(\cancel{r + u})}{(\cancel{r + u})} - \cfrac{3(\cancel{r + u})}{(\cancel{r + u})}\)
\(= 4s + 3\)
We then combine both binomials to get the factored expression:
Therefore, we can determine that \(4s(r + u) - 3(r + u)\) fully factored is \(\boldsymbol{(4s + 3)(r + u)}\).
\(2y(x - 3) + 4z(3 - x)\)
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 \(-1\):
\(2y(x - 3) - 4z(x - 3)\)
\(\text{GFC} = (x - 3)\)
Next, we need to divide the GCF out of every term in the polynomial. This is how we get the second binomial:
\(= \cfrac{2y(\cancel{x - 3})}{(\cancel{x - 3})} - \cfrac{4z(\cancel{x - 3})}{\cancel{x - 3}}\)
\(= 2y - 4z\)
We then combine both binomials to get the factored expression:
Therefore, we can determine that \(2y(x - 3) + 4z(3 - x)\) fully factored is \(\boldsymbol{(2y - 4z)(x - 3)}\).
This is commonly done when there are pairs of terms that can be factored together.
The process for factoring by grouping is:
Fully factor \(ax + ay + 3x + 3y\)
First, we can group the first and last sets of terms into separate brackets:
Next, we can factor out the GFC from each separate binomial:
Finally, we can factor out the common binomial (in this case \(x + y\):
Therefore, we can determine that \(ax + ay + 3x + 3y\) fully factored is \(\boldsymbol{(x + y)(a + 3)}\).
\(4x^2 + 6xy + 12y + 8x\)
First, we can group the first and last pairs of terms into separate brackets:
Next, we can factor out the GFC from each separate binomial:
Finally, we can factor out the common binomial (in this case \(2x + 3y\):
Therefore, we can determine that \(4x^2 + 6xy + 12y + 8x\) fully factored is \(\boldsymbol{(2x + 3y)(2x + 4)}\).
\(y^2 + 3y - ay - 3a\)
First, we can group the first and last pairs of terms into separate brackets:
Next, we can factor out the GFC from each separate binomial:
Finally, we can factor out the common binomial (in this case \(y + 3\):
Therefore, we can determine that \(y^2 + 3y - ay - 3a\) fully factored is \(\boldsymbol{(y + 3)(y - a)}\).
The formula for the surface area of a rectangular prism is \(SA = 2lw + 2lh + 2wh\).
i. All we need to do is factor out the GFC (in this case \(2\)):
Therefore, we can determine that the formula in factored form is \(\boldsymbol{SA = 2(lw + lh + wh)}\).
ii. First, we can plug the values into the original formula:
\(SA = 2lw + 2lh + 2wh\)
\(SA = 2(10)(5) + 2(10)(8) + 2(5)(8)\)
\(SA = 100 + 160 + 80\)
\(SA = 340\;[\text{cm}^2]\)
Next, we can plug the values into the factored formula:
\(SA = 2(lw + lh + wh)\)
\(SA = 2((10)(5) + (10)(8) + (5)(8))\)
\(SA = 2(50 + 80 + 40)\)
\(SA = 2(170)\)
\(SA = 340\;[\text{cm}^2]\)
Therefore, we can determine that the Surface Area in both the original formula and the factored formula is \(\boldsymbol{340\;[\textbf{cm}²]}\). This is because the formulas are equivalent regardless of their format.