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MA158 — Spring 2026
The following formulas are essential tools for factoring and simplifying algebraic expressions:
Difference of Squares:
\[x^2 - y^2 = (x+y)(x-y)\]Perfect Square Trinomials:
\[(x+y)^2 = x^2 + 2xy + y^2\] \[(x-y)^2 = x^2 - 2xy + y^2\]Sum and Difference of Cubes:
\[x^3 - y^3 = (x-y)(x^2 + xy + y^2)\] \[x^3 + y^3 = (x+y)(x^2 - xy + y^2)\]Important Note: The expressions \(x^2 + xy + y^2\) and \(x^2 - xy + y^2\) that appear in the cube formulas are usually irreducible. In most cases, these cannot be factored further over the real numbers.
When factoring or simplifying expressions with fractional or negative exponents, always factor out the term raised to the smallest power when taking out the greatest common factor (GCF).
Factor: \((x+6)^{-2}(x+4) + (x+6)^{-3}(2x+11)\)
Initial attempt (incorrect approach):
If we tried to factor out \((x+6)^{-2}\):
\[= (x+6)^{-2}\left[(x+4) + (x+6)^{-1}(2x+11)\right]\]Note: This approach creates a problem because we end up with \(-3 - (-2) = -1\), but we want \(-3\) because \(-3 < -2\).
Correct approach — Let's restart:
We need to factor out \((x+6)^{-3}\) (the smallest power):
\[\begin{align} (x+6)^{-2}(x+4) + (x+6)^{-3}(2x+11) &= (x+6)^{-2}(x+6)^{-1}(x+6)^1(x+4) + (x+6)^{-3}(2x+11)\\ &= (x+6)^{-3}(x+6)(x+4) + (x+6)^{-3}(2x+11)\\ &= (x+6)^{-3}\left[(x+6)(x+4) + (2x+11)\right]\\ &= (x+6)^{-3}\left[x^2 + 6x + 4x + 24 + 2x + 11\right]\\ &= (x+6)^{-3}\left[x^2 + 12x + 35\right] \end{align}\]Final Answer: \((x+6)^{-3}(x^2 + 12x + 35)\)
Factor: \((2x^3 + 6x^2 - 8x - 24)^{-1}\)
Step 1: Factor by grouping
First, we work with the polynomial in the base:
\[\begin{align} 2x^3 + 6x^2 - 8x - 24 &= (2x^3 + 6x^2) - (8x + 24)\\ &= 2x^2(x+3) - 8(x+3)\\ &= (2x^2 - 8)(x+3)\\ &= 2(x^2 - 4)(x+3) \end{align}\]Step 2: Apply difference of squares
Now we can factor \(x^2 - 4\) further:
\[= 2(x-2)(x+2)(x+3)\]Step 3: Apply the negative exponent to the factored form
\[\begin{align} \left(2x^3 + 6x^2 - 8x - 24\right)^{-1} &= \left(2(x^2-4)(x+3)\right)^{-1}\\ &= \left(2(x-2)(x+2)(x+3)\right)^{-1}\\ &= 2^{-1}(x-2)^{-1}(x+2)^{-1}(x+3)^{-1} \end{align}\]Final Answer: \(\displaystyle\frac{1}{2(x-2)(x+2)(x+3)}\) or \(2^{-1}(x-2)^{-1}(x+2)^{-1}(x+3)^{-1}\)
When simplifying algebraic expressions, follow this general approach:
1. Always check for a GCF first — Factor out the greatest common factor before attempting other methods. For negative or fractional exponents, factor out the term with the smallest power.
2. Look for special patterns — Check if the expression matches difference of squares, perfect square trinomials, or sum/difference of cubes.
3. Try factoring by grouping — For polynomials with four or more terms, group terms strategically to reveal common factors.
4. Use the ac-method for quadratics — For expressions of the form \(ax^2 + bx + c\), find two numbers that multiply to \(ac\) and add to \(b\).
5. Check your work — You can always verify factoring by expanding your answer to see if you get back to the original expression.
Factoring and expanding are inverse operations. Expanding breaks down parentheses and combines like terms, while factoring groups terms and creates parentheses. Being comfortable with both processes is essential for simplifying algebraic expressions effectively.