6.4: The Greatest Common Factor
Factoring Method
In the last two types of problems, we knew one of the factors and were able to determine the other factor through division. Suppose, now, we’re given the product without any factors. Our problem is to find the factors, if possible. This procedure and the previous two procedures are based on the distributive property.
We will use the distributive property in reverse.
\(\underbrace{a b+a c}_{\text {product }}=\underbrace{a(b+c)}_{\text {factors }}\)
We notice that in the product, \(a\) is common to both terms. (In fact, \(a\) is a common factor of both terms.) Since \(a\) is common to both terms, we will factor it out and write
\(a ()\)
Now we need to determine what to place inside the parentheses. This is the procedure of the previous section. Divide each term of the product by the known factor \(a\).
\(\dfrac{ab}{a} = b\) and \(\dfrac{ac}{a} = c\)
Thus, \(b\) and \(c\) are the required terms of the other factor. Hence,
\(ab + ac = a(b+c)\)
When factoring a monomial from a polynomial, we seek out factors that are not only common to each term of the polynomial but factors that have these properties:
- The numerical coefficients are the largest common numerical coefficients.
- The variables possess the largest exponents common to all the variables.
Greatest Common Factor
A monomial factor that meets the above two requirements is called the greatest common factor of the polynomial.
Sample Set A
Factor \(3x - 18\)
The greatest common factor is \(3\).
\(\begin{array}{flushleft}
3x-18&=&3 \cdot x - 3 \cdot 6&\text{Factor out } 3\\
3x-18&=&3()&\text{Divide each term of the product by } 3\\
&&&\dfrac{3x}{3} = x \text{ and } \dfrac{-18}{3} = -6\\
&&&\text{(Try to perform this division mentally.}\\
3x-18&=&3(x-6)
\end{array}\)
Factor \(9x^3+18x^2+27x\)
Notice that \(9x\) is the greatest common factor.
\(9x^3 + 18x^2 + 27x = 9x \cdot x^2 + 9x \cdot 2x + 9x \cdot 3\). Factor out \(9x\)
\(9x^3 + 18x^2 + 27x = 9x()\) Mentally divide \(9x\) into each term of the product\)
\(9x^3 + 18x^2 + 27x = 9x(x^2+2x+3)\)
Factor \(10x^2y^3 - 20xy^4 - 35y^5\).
Notice that \(5y^3\) is the greatest common factor. Factor out \(5y^3\).
\(10x^2y^3-20xy^4-35y^5 = 5y^3()\)
Mentally divide \(5y^3\) into each term of the product and place the resulting quotients inside the ( ).
\(10x^2y^3-20xy^4-35y^5=5y^3(2x^2-4xy-7y^2)\)
Factor \(-12x^5 + 8x^3 - 4x^2\).
We see that the greatest common factor is \(-4x^2\).
\(-12x^5 + 8x^3 - 4x^2 = -4x^2()\)
Mentally dividing \(-4x^2\) into each term of the product, we get
\(-12x^5 + 8x^3 - 4x^2 = -4x^2(3x^3 - 2x + 1\)
Practice Set A
Factor \(4x−48\).
- Answer
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\(4(x−12)\)
Factor \(6y^3 + 24y^2 + 36y\)
- Answer
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\(6y(y^2 + 4y + 6)\)
Factor \(10a^5b^4 - 14a^4b^5-8b^6\)
- Answer
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\(2b^4(5a^5 - 7a^4b - 4b^2\)
Factor \(-14m^4 + 28m^2 - 7m\)
- Answer
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\(-7m(2m^2 - 4m + 1\)
Consider this problem: factor \(Ax+Ay\). Surely, \(Ax+Ay=A(x+y)\). We know from the very beginning of our study of algebra that letters represent single quantities. We also know that a quantity occurring within a set of parentheses is to be considered as a single quantity. Suppose that the letter \(A\) is representing the quantity \((a+b)\). Then we have
\(Ax + Ay = A(x+y)\)
\((a+b)x + (a+b)y = (a+b)(x+y)\)
When we observe the expression
\((a+b)x + (a+b)y\)
we notice that \((a+b)\) is common to both terms. Since it is common, we factor it out.
\((a+b)( )\)
As usual, we determine what to place inside the parentheses by dividing each term of the product by \((a+b)\).
\(\dfrac{(a+b)x}{(a+b)} = x\) and \(\dfrac{(a+b)y}{(a+b)} = y\)
This is a forerunner of the factoring that will be done in Section 5.4.
Sample Set B
Factor \((x−7)a+(x−7)b\).
Notice that \((x-7)\) is the greatest common factor. Factor out \((x-7)\).
\((x-7)a + (x-7)b = (x-7)( )\)
Then, \(\dfrac{(x-7)a}{(x-y)} = a \text{ and } \dfrac{(x-7)b}{(x-7)} = b\)
\((x-7)a+(x-7)b = (x-7)(a+b)\)
Factor \(3x^2(x+1)-5x(x+1)\).
Notice that \(x\) and \((x+1)\) are common to both terms. Factor them out. We'll perform this factorization by letting \(A = x(x+1)\). Then we have
\(3xA-5A = A(3x-5)\)
But \(A = x(x+1)\), so
\(3x^2(x+1)-5x(x+1) = x(x+1)(3x-5)\)
Practice Set B
Factor \((y+4)a+(y+4)b\).
- Answer
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\((y+4)(a+b)\)
Factor \(8m^3(n-4) - 6m^2(n-4)\)
- Answer
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\(2m^2(n-4)(4m-3)\)
Exercises
For the following problems, factor the polynomials.
\(9a+18\)
- Answer
-
\(9(a+2)\)
\(6a+24\)
\(8b+12\)
- Answer
-
\(4(2b+3)\)
\(16x+12\)
\(4x−6\)
- Answer
-
\(2(2x−3)\)
\(8x−14\)
\(21y−28\)
- Answer
-
\(7(3y−4)\)
\(16f−36\)
\(12x^2 + 18x\)
- Answer
-
\(6x(2x+3)\)
\(10y^2 + 15y\)
\(8y^2 + 18\)
- Answer
-
\(2(4y^2 + 9)\)
\(7x^2 - 21\)
\(3y^2 - 6\)
- Answer
-
\(3(y^2 - 2)\)
\(2x^2-2\)
\(6y^2-6y\)
- Answer
-
\(6y(y−1)\)
\(ax^2-a\)
\(by^2 + b\)
- Answer
-
\(b(y^2 + 1)\)
\(7by^2 + 14b\)
\(5a^2x^2 + 10x\)
- Answer
-
\(5x(a^2x + 2)\)
\(24ax^2 + 28a\)
\(10x^2 + 5x - 15\)
- Answer
-
\(5(2x^2 + x - 3)\)
\(12x^2 - 8x - 16\)
\(15y^3 - 24y + 9\)
- Answer
-
\(3(5y^3 - 8y + 3)\)
\(ax^2 + ax + a\)
\(by^3 + by^2 + by + b\)
- Answer
-
\(b(y^3 + y^2 + y + 1)\)
\(2y^2 + 6y + 4xy\)
\(9x^2 + 6xy + 4x\)
- Answer
-
\(x(9x+6y+4)\)
\(30a^2b^2 + 40a^2b^2 + 50a^2b^2\)
\(13x^2y^5c - 26x^2y^5c - 39x^2y^5\)
- Answer
-
\(13x^2y^5(-c-3)\)
\(-4x^2-12x-8\)
\(-6y^3 - 8y^2 - 14y + 10\)
- Answer
-
\(-2(3y^3 + 4y^2 + 7y - 5)\)
\(Ab+Ac\)
\(Nx+Ny\)
- Answer
-
\(N(x+y)\)
\(Qx+Qy\)
\(Ax−Ay\)
- Answer
-
\(A(x−y)\)
\((x+4)b+(x+4)c\)
\((x−9)a+(x−9)b\)
- Answer
-
\((x−9)(a+b)\)
\((2x+7)a+(2x+7)b\)
\((9a−b)w−(9a−b)x\)
- Answer
-
\((9a−b)(w−x)\)
\((5−v)X+(5−v)Y\)
\(3x^5y^4 - 12x^3y^4 + 27x^5y^3 - 6x^2y^6\)
- Answer
-
\(3x^2y^3(x^3y - 4xy + 9x^3-2y^3)\)
\(8a^3b^{15} + 24a^2b^{14} + 48a^3b^6 - 20a^3b^7 + 80a^4b^6 - 4a^3b^6 - 4a^3b^7 + 4a^2b\)
\(-8x^3y^2 - 3x^3y^2 + 16x^4y^3 + 2x^2y\)
- Answer
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\(-x^2y(11xy-16x^2y^2-2)\)
Exercises for Review
A quantity plus \(21\)% more of that quantity is \(26.25\). What is the original quantity?
Solve the equation \(6(t−1)=4(5−s)\) if \(s=2\).
- Answer
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\(t=3\)
Given that \(4a^3\) is a factor of \(8a^3 - 12a^2\), find the other factor.