Factoring a polynomial is to write it as the product of simpler polynomials. How to factor a polynomial using a common factor? Questions are presented along with detailed solutions and explanations are included.
It is a factorization method based on the law of distributivity which is
\[
\Large{\color{red} {a(b + c) = a \cdot b + a \cdot c}}
\]
and used in reverse as follows
\[
\Large{\color{red} { a \cdot b + a \cdot c = a(b + c) }}
\]
\( a \) is a common factor to \( a \cdot b \) and \( a \cdot c \) is therefore factored out.
\( 3x^2 - x = 3x \cdot \textcolor{red}{x} - 1 \cdot \textcolor{red}{x} = \textcolor{red}{x}(3x - 1) \quad \) The common factor is \(x\).
NOTE: It is very easy to check if your factorization is correct by expanding the factored form to see if you get the original polynomial.
Example: Check that \( 3x^2 - x = x(3x - 1) \) by expanding \( x(3x - 1) \quad \) using the distributive property:
\( x(3x - 1) = x \cdot 3x + x \cdot (-1) = 3x^2 - x \quad \), which matches the original expression.
a) \( 9x - 6 \)
b) \( x^2 - x \)
c) \( 3x + 12xy \)
d) \( 16x^3 + 8x^2y + 4xy^2 \)
e) \( 2x^4(x + 5) + x^2(x + 5) \)
a)
Find any common factors in the two terms of the binomial expression \( 9x - 6 \) by expressing both terms \( 9x \) and \( 6 \) as prime factorizations.
Using prime factorization: \[ 9x - 6 = \color{red}{3} \cdot 3 \cdot x - 2 \cdot \color{red}{3} \]The greatest common factor (GCF) is \( \color{red}{3} \) We factor it out from the expression. \[ 9x - 6 = \color{red}{3}(3x - 2) \] This is the simplified, factored form of the binomial using its greatest common factor.
b)
The prime factorization of \( x^2 \) and \( x \) is needed to find the greatest common factor in \( x^2 - x \). \[ x^2 - x = \color{red}{x} \cdot x - \color{red}{x} = \color{red}{x} \cdot x - 1 \cdot \color{red}{x} \] The greatest common factor is \( x \), and it is therefore factored out. Hence, \[ x^2 - x = \color{red}{x}(x - 1) \]c)
To factor the expression \( 3x + 12xy \), we begin by finding the prime factorizations of the individual terms.
The term \( 3x \) factors into \( 3 \cdot x \), and the term \( 12xy \) factors into \( 3 \cdot 4 \cdot x \cdot y \).
So we can rewrite the expression as: \[ 3x + 12xy = 3 \cdot x + 3 \cdot 4 \cdot x \cdot y \] We observe that both terms have a common factor of \( \color{red} {3x} \). Factoring out \( 3x \), we get: \[ 3x + 12xy = {\color{red}{3x}} (1 + 4y) \] Thus, the greatest common factor is \( 3x \), and the factored form of the expression is \( 3x(1 + 4y) \).
d)
To find the greatest common factor (GCF) of the expression \( 16x^3 + 8x^2y + 4xy^2 \), we first determine the prime factorization of each term:
The greatest common factor (GCF) among these terms is: \( 2 \cdot 2 \cdot x = 4x \)
Now factor out the GCF from the expression: \[ 16x^3 + 8x^2y + 4xy^2 = \] \[ 4x \left( 2 \cdot 2 \cdot x \cdot x + 2 \cdot x \cdot y + y \cdot y \right) \] \[ = 4x \left( 4x^2 + 2xy + y^2 \right) \]
e)
We note that \(x + 5\) is a common factor which can be factored out as follows:
\[ 2x^4(x + 5) + x^2(x + 5) = (x + 5)(2x^4 + x^2) \]We now find the greatest common factor (GCF) of the terms \(2x^4\) and \(x^2\), and factor it out:
\[ 2x^4 + x^2 = 2 \cdot x \cdot x \cdot x \cdot x + x \cdot x = x^2(2x^2 + 1) \]The complete factoring of the expression \(2x^4(x + 5) + x^2(x + 5)\) is:
\[ 2x^4(x + 5) + x^2(x + 5) = x^2(x + 5)(2x^2 + 1) \]
Use common factors to factor completely the following polynomials:
a) \( -3x + 9 \)
b) \( 28x + 2x^2 \)
c) \( 11xy + 55x^2y \)
d) \( 20xy + 35x^2y - 15xy^2 \)
e) \( 5y(x + 1) + 10y^2(x + 1) - 15xy(x + 1) \)
a)
Find any common factors in the two terms of \( -3x + 9 \) by expressing both terms \( 3x \) and \( 9 \) in the given binomial as prime factorization.
\[ -3x + 9 = -\color{red}{3} \cdot x + \color{red}{3} \cdot 3 \] The greatest common factor is \( \color{red}{3} \) and is factored out. Hence \[ -3x + 9 = \color{red}{3}(-x + 3) = -3(x - 3) \]
b)
Write the prime factorization of each of the terms in the given polynomial \( 28x + 2x^2 \).
\[ 28x + 2x^2 = \color{red}{2} \cdot 2 \cdot 7 \cdot \color{red}{x} + \color{red}{2} \cdot \color{red}{x} \cdot x \] The greatest common factor is \( \color{red}{2x} \) and is factored out. Hence \[ 28x + 2x^2 = \color{red}{2x}(14 + x) \]
c)
Write the prime factorization of each of the terms in the given polynomial \( 11xy + 55x^2y \).
\[ 11xy + 55x^2y = \color{red}{11} \cdot x \cdot y + 5 \cdot \color{red}{11} \cdot x \cdot x \cdot \color{red}{y} \] The greatest common factor is \( \color{red}{11xy} \) and is factored out. Hence, \[ 11xy + 55x^2y = \color{red}{11xy}(1 + 5x) \]
d)
Write the prime factorization of each of the terms in the given polynomial \( 20xy + 35x^2y - 15xy^2 \).
\[ 20xy + 35x^2y - 15xy^2 = 2 \cdot 2 \cdot \color{red}{5} \cdot \color{red}{x} \cdot \color{red}{y} + \color{red}{5} \cdot 7 \cdot \color{red}{x} \cdot x \cdot \color{red}{y} - 3 \cdot \color{red}{5} \cdot \color{red}{x} \cdot \color{red}{y} \cdot y \] The greatest common factor is \( \color{red}{5xy} \) and is factored out. Hence \[ 20xy + 35x^2y - 15xy^2 = \color{red}{5xy}(4 + 7x - 3y) \]
e)
We start by factoring out the common factor \( (x + 1) \) in the given polynomial.
\[ 5y(x + 1) + 10y^2(x + 1) - 15xy(x + 1) = (x + 1)(5y + 10y^2 - 15xy) \] We now factor the polynomial \( 5y + 10y^2 - 15xy \) using the GCF of all three terms. \[ 5y + 10y^2 - 15xy = {5 \cdot y} + 2 \cdot {5 \cdot y} \cdot y - 3 \cdot {5 \cdot y} \cdot x = \color{red}{5 \cdot y}(1 + 2y - 3xy) \] The given polynomial may be factored as follows. \[ 5y(x + 1) + 10y^2(x + 1) - 15xy(x + 1) = \color{red} {5y(x + 1)(1 + 2y - 3x)} \]