Factoring by Adding and Subtracting Terms
When an expression cannot be factored using standard methods (such as grouping or trinomial factoring), we can sometimes create a recognizable pattern—like a perfect square trinomial or a difference of squares—by adding and subtracting the same term. This preserves the value of the original expression while changing its form.
Method and Strategy:
1. Identify the terms that look like squares (e.g., $x^4$ and $1$).
2. Determine what middle term is required to make them a perfect square trinomial $(a^2 \pm 2ab + b^2)$.
3. Add and subtract that necessary term so that the overall value remains unchanged.
4. Group the perfect square together and rewrite the expression as a difference of squares:
\[ A^2 – B^2 = (A – B)(A + B) \]
Example:
Let’s factor the expression $x^4 + x^2y^2 + y^4$. By adding and subtracting $x^2y^2$, we get:
\[
x^4 + x^2y^2 + y^4 = x^4 + 2x^2y^2 + y^4 – x^2y^2
\]
\[
= (x^2 + y^2)^2 – (xy)^2
\]
\[
= (x^2 + y^2 – xy)(x^2 + y^2 + xy)
\]
QUESTION 19
Which of the following is a factor of the expression $x^4 + x^2 + 1$?
\[
\text{A) } x+1 \quad
\text{B) } x-1 \quad
\text{C) } x^2+1 \quad
\text{D) } x^2+x+1 \quad
\text{E) } x^2-x-1
\]
Solution:
To rewrite the expression as a perfect square trinomial, we add and subtract $x^2$:
\[
x^4 + x^2 + 1 = x^4 + 2x^2 + 1 – x^2
\]
\[
= (x^2+1)^2 – x^2
\]
\[
= (x^2 + 1 – x)(x^2 + 1 + x)
\]
\[
= (x^2 – x + 1)(x^2 + x + 1)
\]
\(\textbf{Correct Answer: D} \)
QUESTION 20
Which of the following is a factor of the expression $a^2 – b^2 + 2a – 4b – 3$?
\[
\text{A) } a-b-3 \quad
\text{B) } a+b+2 \quad
\text{C) } a-b+2 \quad
\text{D) } a+b+1 \quad
\text{E) } a-b-1
\]
Solution:
By adding and subtracting $1$:
\[
a^2 – b^2 + 2a – 4b – 3 = a^2 – b^2 + 2a – 4b – 3 + 1 – 1
\]
\[
= a^2 + 2a + 1 – (b^2 + 4b + 4)
\]
\[
= (a+1)^2 – (b+2)^2
\]
Using the difference of squares formula:
\[
= (a+1 – (b+2))(a+1 + b + 2)
\]
\[
= (a – b – 1)(a + b + 3)
\]
\(\textbf{Correct Answer: E} \)
Simplification of Rational Expressions
To simplify rational expressions of the form $\displaystyle\frac{A(x)}{B(x)}$ where $B(x) \neq 0$, factor both the numerator and the denominator completely. Then, cancel out the common factors.
QUESTION 21
What is the simplest form of the following expression?
\[
\frac{2x^3 + 16}{(x^2 – 4)(x^2 – 2x + 4)}
\]
\[
\text{A) } 2 \quad
\text{B) } \frac{2}{x-2} \quad
\text{C) } \frac{2}{x+2} \quad
\text{D) } \frac{1}{x} \quad
\text{E) } x
\]
Solution:
\[
\frac{2x^3 + 16}{(x^2 – 4)(x^2 – 2x + 4)}
= \frac{2(x^3 + 2^3)}{(x-2)(x+2)(x^2 – 2x + 4)}
\]
Using the sum of cubes factorization $x^3 + 2^3 = (x+2)(x^2 – 2x + 4)$:
\[
= \frac{2(x+2)(x^2 – 2x + 4)}{(x-2)(x+2)(x^2 – 2x + 4)}
\]
\[
= \frac{2}{x-2}
\]
\(\textbf{Correct Answer: B} \)
QUESTION 22
What is the simplified form of the following expression?
\[
\left( \frac{1}{x-1} + x^2 + x + 1 \right) : \frac{x^2}{x-1}
\]
\[
\text{A) } x^2 \quad
\text{B) } \frac{1}{x-1} \quad
\text{C) } x-1\quad
\text{D) } \frac{1}{x} \quad
\text{E) } x
\]
Solution:
\[
\left( \frac{1}{x-1} + x^2 + x + 1 \right) \div \frac{x^2}{x-1}
\]
Find a common denominator inside the parentheses:
\[
= \left( \frac{1 + (x-1)(x^2 + x + 1)}{x-1} \right) \cdot \frac{x-1}{x^2}
\]
Recognize the difference of cubes $(x-1)(x^2 + x + 1) = x^3 – 1$:
\[
= \frac{1 + x^3 – 1}{x-1} \cdot \frac{x-1}{x^2} = \frac{x^3}{x^2} = x
\]
\(\textbf{Correct Answer: E} \)
QUESTION 23
What is the result of the following operation?
\[
\frac{2x^2}{x-1} + \frac{5x-3}{1-x}
\]
\[
\text{A) } 2x+3 \quad
\text{B) } 2x-3 \quad
\text{C) } x-3\quad
\text{D) } x+3 \quad
\text{E) } x+1
\]
Solution:
\[
\frac{2x^2}{x-1} + \frac{5x-3}{1-x} = \frac{2x^2}{x-1} + \frac{5x-3}{-(x-1)}
\]
\[
= \frac{2x^2}{x-1} – \frac{5x-3}{x-1}
\]
\[
= \frac{2x^2 – 5x + 3}{x-1}
\]
Factor the numerator:
\[
= \frac{(2x-3)(x-1)}{x-1} = 2x – 3
\]
\(\textbf{Correct Answer: B} \)
QUESTION 24
What is the result of the following operation?
\[
\frac{b^2 x^2 – 4}{b} \cdot \frac{ax-1}{abx^2 + (2a – b)x -2} + \frac{2}{b}
\]
\[
\text{A) } 1 \quad
\text{B) } -1 \quad
\text{C) } x\quad
\text{D) } -x \quad
\text{E) } b
\]
Solution:
Factor the components of the first expression:
\[
\frac{b^2 x^2 – 4}{b} \cdot \frac{ax – 1}{abx^2 + (2a – b)x – 2} + \frac{2}{b}
\]
\[
= \frac{(bx – 2)(bx + 2)}{b} \cdot \frac{ax – 1}{(ax – 1)(bx + 2)} + \frac{2}{b}
\]
Cancel out common terms:
\[
= \frac{bx – 2}{b} + \frac{2}{b} = \frac{bx – 2 + 2}{b}
\]
\[
= \frac{bx}{b} = x
\]
\(\textbf{Correct Answer: C} \)
QUESTION 25
What is the result of the following operation?
\[
\frac{x – \frac{1}{x}}{x+1} : \frac{x – \frac{1}{x^2}}{x^2 + x + 1}
\]
\[
\text{A) } 1 \quad
\text{B) } x \quad
\text{C) } x^2\quad
\text{D) } x^3 \quad
\text{E) } x^4
\]
Solution:
Simplify fractions in numerators:
\[
\frac{x – \frac{1}{x}}{x+1} \div \frac{x – \frac{1}{x^2}}{x^2 + x + 1}
\]
\[
= \frac{x^2 – 1}{x(x+1)} \div \frac{x^3 – 1}{x^2(x^2 + x + 1)}
\]
Invert and multiply, then factor completely:
\[
= \frac{(x-1)(x+1)}{x(x+1)} \cdot \frac{x^2 (x^2 + x + 1)}{(x-1)(x^2 + x + 1)}
\]
\[
= \frac{x^2}{x} = x
\]
\(\textbf{Correct Answer: B} \)
QUESTION 26
What is the result of the following operation?
\[
\frac{x \sqrt{y} – y \sqrt{x}}{\sqrt{x} – \sqrt{y}} + \frac{1 – xy}{\sqrt{xy} – 1}
\]
\[
\text{A) } 1 \quad
\text{B) } -1 \quad
\text{C) } 1-x\quad
\text{D) } x-1 \quad
\text{E) } y
\]
Solution:
Factor out common roots in the first numerator, and factor the second numerator using difference of squares:
\[
\frac{x \sqrt{y} – y \sqrt{x}}{\sqrt{x} – \sqrt{y}} + \frac{1 – xy}{\sqrt{xy} – 1}
\]
\[
= \frac{\sqrt{xy} (\sqrt{x} – \sqrt{y})}{\sqrt{x} – \sqrt{y}} + \frac{(1 – \sqrt{xy})(1 + \sqrt{xy})}{-(1 – \sqrt{xy})}
\]
\[
= \sqrt{xy} – (1 + \sqrt{xy}) = -1
\]
\(\textbf{Correct Answer: B} \)
QUESTION 27
Given that:
\[
\frac{4^x + 2^{x+y+1} + 4^y}{4^x – 4^y} = \frac{3}{2}
\]
What is the value of $2^{x-y}$?
\[
\text{A) } 1 \quad
\text{B) } 2 \quad
\text{C) } 3\quad
\text{D) } 4 \quad
\text{E) } 5
\]
Solution:
\[
\frac{4^x + 2^{x+y+1} + 4^y}{4^x – 4^y} = \frac{3}{2}
\]
Rewrite with base 2:
\[
\Rightarrow \frac{(2^x)^2 + 2 \cdot 2^x \cdot 2^y + (2^y)^2}{(2^x)^2 – (2^y)^2} = \frac{3}{2}
\]
Factor as a perfect square and a difference of squares:
\[
\Rightarrow \frac{(2^x + 2^y)^2}{(2^x – 2^y)(2^x + 2^y)} = \frac{3}{2}
\]
\[
\Rightarrow \frac{2^x + 2^y}{2^x – 2^y} = \frac{3}{2}
\]
Cross-multiply:
\[
\Rightarrow 2 \cdot 2^x + 2 \cdot 2^y = 3 \cdot 2^x – 3 \cdot 2^y
\]
\[
\Rightarrow 5 \cdot 2^y = 2^x
\]
Divide by $2^y$:
\[
\Rightarrow 5 = \frac{2^x}{2^y} \Rightarrow 2^{x-y} = 5
\]
\(\textbf{Correct Answer: E} \)
Completing the Square for the Quadratic Trinomial $Ax^2 + Bx + C$
1) When $A = 1$
A perfect square is created by adding and subtracting $\left(\frac{B}{2}\right)^2$:
\[
x^2 + Bx + C = x^2 + Bx + C + \left(\frac{B}{2}\right)^2 – \left(\frac{B}{2}\right)^2
\]
\[
= \left(x + \frac{B}{2} \right)^2 + C – \frac{B^2}{4}
\]
Example:
Let’s complete the square for $x^2 + x + 1$.
Since $B = 1$, we add and subtract $\left(\frac{1}{2}\right)^2 = \frac{1}{4}$:
\[
x^2 + x + 1 = x^2 + x + 1 + \frac{1}{4} – \frac{1}{4}
\]
\[
= \left(x + \frac{1}{2} \right)^2 + 1 – \frac{1}{4}
\]
\[
= \left(x + \frac{1}{2} \right)^2 + \frac{3}{4}
\]
2) When $A \neq 1$
\[
Ax^2 + Bx + C = A \left(x^2 + \frac{B}{A} x + \frac{C}{A} \right)
\]
Factor out $A$ first, then complete the square inside the parentheses.
Example:
Let’s complete the square for $2x^2 – 3x + 4$.
\[
2x^2 – 3x + 4 = 2 \left(x^2 – \frac{3}{2} x + 2\right)
\]
Inside the parentheses, the coefficient of the linear term is $-\frac{3}{2}$. We add and subtract $\left(-\frac{3}{2} \cdot \frac{1}{2}\right)^2 = \frac{9}{16}$:
\[
2 \left(x^2 – \frac{3}{2} x + 2 \right) = 2 \left(x^2 – \frac{3}{2} x + \frac{9}{16} + 2 – \frac{9}{16} \right)
\]
\[
= 2 \left(\left(x – \frac{3}{4}\right)^2 + \frac{23}{16} \right)
\]
\[
= 2 \left(x – \frac{3}{4}\right)^2 + \frac{23}{8}
\]
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