Factoring by Grouping and GCF Extraction
When factoring an algebraic expression, if a greatest common factor (GCF) exists across all terms, the expression is factored by extracting this common term.
In cases where a common factor is not immediately shared by every single term, the expression can be factored by grouping appropriate terms together and extracting the common factors within those individual groups.
Examples:
\( \bullet \quad 6xy + 3xz \;- \;27xt = 3x (2y + z \;-\; 9t) \)
\(\bullet \quad -3x^3 – 6x^2 + 81x = -3x (x^2 + 2x – 27) \)
\(\bullet \quad x^3 y^2 – x^2 y + xy = xy (x^2 y – x + 1) \)
\(\bullet \quad \sqrt{a} + a + a\sqrt{a} = \sqrt{a} (1 + \sqrt{a} + a) \)
\(\bullet \quad 2^x – 2^{2x} + 2^{3x} = 2^x (1 – 2^x + 2^{2x}) \)
\(\bullet \quad 3^{66} + 3^{99} = 3^{66} (1 + 3^{33}) \)
\(\bullet \quad x^{-96} + x^{-95} = x^{-96} (1 + x) \)
\(\bullet \quad ax + bx – ay – by = x (a + b) – y (a + b) \)
\[
= (a + b) (x – y)
\]
\(\bullet \quad 21xy + 7x + 9y + 3 = 7x (3y + 1) + 3 (3y + 1) \)
\[
= (3y + 1) (7x + 3)
\]
\(\bullet \quad a^3 + 2a^2 + a + 2 = a^2 (a + 2) + (a + 2) \)
\[
= (a + 2) (a^2 + 1)
\]
\(\bullet \quad (a – b)^2 (a – c) + (b – a) (a – c)^2 \)
\[
= (a – b)^2 (a – c) – (a – b) (a – c)^2
\]
\[
= (a – b) (a – c) (a – b – (a – c))
\]
\[
= (a – b) (a – c) (c – b)
\]
\(\bullet \quad -\sqrt{xy} + \sqrt{y} + x – \sqrt{x} \)
\[
= -\sqrt{y} (\sqrt{x} – 1) + \sqrt{x} (\sqrt{x} – 1)
\]
\[
= (\sqrt{x} – 1) (\sqrt{x} – \sqrt{y})
\]
QUESTION 1
Which of the following is not a factor of the algebraic expression
\[
xy (x + 3)^2 – x^2 y (x + 3) + xy (x + 3)
\]
\[
\text{A) } x \quad
\text{B) } y \quad
\text{C) } 3 \quad
\text{D) } 4 \quad
\text{E) } x+3
\]
Solution:
\[
xy (x + 3)^2 – x^2 y (x + 3) + xy (x + 3)
\]
Factoring out the greatest common factor \( xy(x + 3) \):
\[
= xy (x + 3) (x + 3 – x + 1)
\]
\[
= 4xy (x + 3)
\]
Evaluating the components reveals that 3 is not a factor of the expression.
\(\textbf{Correct Answer: C} \)
QUESTION 2
Which of the following choices represents a factor of the expression
\[
ab + a^2 b + a^3 b + a^4 b
\]
\[
\text{A) } a^2b \quad
\text{B) } ab^2\quad
\text{C) } a+b \quad
\text{D) } 1+a^2 \quad
\text{E) } 1+a+a^2
\]
Solution:
First, extract the greatest common factor \( ab \):
\[
ab + a^2 b + a^3 b + a^4 b = ab (1 + a + a^2 + a^3)
\]
Next, apply factoring by grouping within the parentheses:
\[
= ab ((1 + a) + a^2 (1 + a))
\]
\[
= ab (1 + a) (1 + a^2)
\]
Thus, \( 1 + a^2 \) is a valid factor.
\(\textbf{Correct Answer: D} \)
QUESTION 3
Which of the following choices represents a factor of the expression
\[
3xy – 20ab – 15xb + 4ya
\]
\[
\text{A) } x+a \quad
\text{B) } y-5b\quad
\text{C) } y+a \quad
\text{D) } y-b \quad
\text{E) } y+b
\]
Solution:
Rearrange the terms to group common variables together:
\[
3xy \;- \; 20ab\; -\; 15xb + 4ya
\]
\[
= 3xy \;- \;15xb + 4ya \;- \;20ab
\]
Factor by grouping the first two terms and the last two terms:
\[
= 3x (y \;-\; 5b) + 4a (y \;- \;5b)
\]
\[
= (y \;- \;5b) (3x + 4a)
\]
Thus, \( y – 5b \) is a factor.
\(\textbf{Correct Answer: B} \)
QUESTION 4
Which of the following choices represents a factor of the expression
\[
2^x + 5^x + 6^x + (15)^x
\]
\[
\text{A) } 1+ 3^x \quad
\text{B) } 2^x + 3^x \quad
\text{C) } 3^x+5^x \quad
\text{D) } 1+ 5^x \quad
\text{E) } 1+2^x
\]
Solution:
Rearrange the expression using the properties of exponents:
\[
2^x + 5^x + 6^x + (15)^x = 2^x + 6^x + 5^x + (15)^x
\]
\[
= 2^x + 2^x \cdot 3^x + 5^x + 5^x \cdot 3^x
\]
Factor by grouping the pairs:
\[
= 2^x (1 + 3^x) + 5^x (1 + 3^x)
\]
\[
= (1 + 3^x) (2^x + 5^x)
\]
Thus, \( 1 + 3^x \) is a factor.
\(\textbf{Correct Answer: A} \)
QUESTION 5
Given that \( a + b = 3 \) and \( b + c = 4 \), determine the numerical value of the expression:
\[
a^2 – bc + ab – ac
\]
\[
\text{A) } 3 \quad
\text{B) } -3 \quad
\text{C) } 2 \quad
\text{D) } -2 \quad
\text{E) } 0
\]
Solution:
Rearrange and factor the expression by grouping:
\[
a^2 – bc + ab – ac = a^2 – ac + ab – bc
\]
\[
= a (a – c) + b (a – c)
\]
\[
= (a – c) (a + b)
\]
We can find the value of \( a – c \) by setting up a system of linear equations and subtracting the second equation from the first:
\[
\begin{array}{c}
a + b = 3 \\
– (b + c = 4)
\end{array}
\]
\[
a – c = -1
\]
Substituting the known values into the factored expression yields:
\[
(a – c)(a + b) = (-1) \cdot 3 = -3
\]
\(\textbf{Correct Answer: B} \)
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