Almost every "hard" problem in the next chapter isn't hard because the calculus idea is hard. It's hard because it hides behind an algebra step that has to happen first: a factoring that cancels a zero denominator, or a conjugate that clears a radical. This section is a fast pass through those moves. They're the ones that show up on repeat starting next chapter.
Every one of these tricks does the same job. Each one turns a \frac{0}{0} mess into something you're allowed to plug a number into. Factoring finds the part of the top and bottom that's secretly the same thing, so it can cancel before you ever divide by zero. Rationalizing does the identical job through a side door: it multiplies by a disguised form of 1, a conjugate over itself. That forces a radical to simplify away where the zero was hiding. Neither trick changes what the function equals. Both just rewrite it into a form that doesn't choke the moment you substitute.
This "survival kit" is a fast pass through three specific algebra moves, chosen because the next chapter leans on exactly these three, on repeat. Factoring a polynomial means rewriting it as a product of simpler pieces so a shared zero cancels: \frac{x^2-9}{x-3} = \frac{(x-3)(x+3)}{x-3} = x+3 for x \neq 3. Rationalizing means multiplying by a conjugate to clear a radical from a numerator or denominator, where a conjugate is an expression that pairs with a radical term to erase it via a difference of squares. Multiplying \frac{\sqrt{x}-2}{x-4} by \frac{\sqrt{x}+2}{\sqrt{x}+2} turns the numerator into x - 4, which then cancels. Simplifying radicals means keeping the exact form rather than a rounded decimal: \sqrt{12} = 2\sqrt{3}, not 3.464\ldots. An exact answer is very often just what the next chapter's problems want.
Formula
Factoring Patterns
Recalled from College Algebra · Polynomials and factoring — first taught there.
Four patterns worth recognizing on sight, each running a multiplication rule in reverse:
Difference of squares: a^2 - b^2 = (a + b)(a - b)
Perfect-square trinomial (both signs): the special products from earlier in this chapter, run in reverse: a^2 + 2ab + b^2 = (a+b)^2, \qquad a^2 - 2ab + b^2 = (a-b)^2
Sum of cubes: a^3 + b^3 = (a+b)(a^2 - ab + b^2)
Difference of cubes: a^3 - b^3 = (a-b)(a^2+ab+b^2)
In both cube patterns, the binomial factor keeps the same sign as the original expression; the trinomial factor always keeps the opposite sign in its middle term, and its last term is always positive.
Formula
Properties of Radicals
Recalled from College Algebra · Exponents and radicals — first taught there.
Valid whenever both sides are defined:
- Product property: \sqrt[n]{ab} = \sqrt[n]{a}\,\sqrt[n]{b}
- Quotient property: \sqrt[n]{\dfrac{a}{b}} = \dfrac{\sqrt[n]{a}}{\sqrt[n]{b}}
- Rational-exponent link: \left(\sqrt[n]{a}\right)^m = \sqrt[n]{a^m} = a^{m/n}
Drag the roots — watch the factored form follow
Drag either root — the factored form stays in sync as you move the roots at x = -3 and x = 2.
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Where does this factored expression change sign?
Click each interval — (x+3)(x-2) is positive outside its roots (x < -3 or x > 2) and negative between them, the same roots as the widget above.
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Worked example
Rationalizing a numerator
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1
Simplify \dfrac{\sqrt{x} - 2}{x - 4} for x \neq 4.
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2conjugate of a - b is a + b
Multiply top and bottom by the conjugate of the numerator, \sqrt{x} + 2. That's a disguised form of 1, so nothing about the value changes: \dfrac{(\sqrt{x}-2)(\sqrt{x}+2)}{(x-4)(\sqrt{x}+2)}.
conjugate of a - b is a + b
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3
The numerator is now a difference of squares: (\sqrt{x})^2 - 2^2 = x - 4.
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4
Now \dfrac{x - 4}{(x-4)(\sqrt{x}+2)}. The (x-4) factor cancels, since x \neq 4: \dfrac{1}{\sqrt{x}+2}.
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Valid step, or invalid?
Sort each algebra claim as a valid step or an invalid one -- two are classic mistakes worth catching now, not mid-calculation next chapter.
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Where this goes next
Every algebraic simplification in the next chapter
Plugging the target value directly into next chapter's expressions almost always gives \frac{0}{0} at the interesting points, the ones worth asking about. Factoring and rationalizing are the two moves that turn that \frac{0}{0} into a form you can substitute into safely. The next chapter uses them on essentially every problem that isn't already in simplified form.
Check your understanding (counts toward your progress)
Factor completely: \,x^2 + 4x - 77
Right on try — credit. Right — the answer had already been shown, so no credit this time. Next try is worth Answer shown after three tries — this one no longer earns credit. You chose to see the answer — this one earns no credit.
Why
Find two numbers that multiply to -77 and add to 4: that's -11 and 7 (check: -11 \times 7 = -77 and -11 + 7 = 4). So x^2 + 4x - 77 = (x + 11)(x - 7).
Why is factoring usually the first move when a limit expression gives \dfrac{0}{0} by direct substitution?
Right on try — credit. Right — the answer had already been shown, so no credit this time. Next try is worth Answer shown after three tries — this one no longer earns credit. You chose to see the answer — this one earns no credit.
Factoring never changes what the expression equals -- it only rewrites it into an equivalent form, valid everywhere except the one canceled point.
\dfrac{0}{0} isn't a fixed value at all -- it's undefined, which is exactly the problem factoring is trying to work around.
Factoring works just as well on standalone expressions as it does inside equations -- there's no such restriction.
Why
A shared factor in the numerator and denominator is what's causing the zero-over-zero in the first place -- cancelling it produces an equivalent expression that no longer chokes at that point.
Practice: Factoring and radicals, limit-ready (3 problems · doesn’t affect your progress)
Simplify \dfrac{\sqrt{x}-3}{x-9} for x\neq9, then evaluate the simplified expression at x=16.
Right on try — credit. Right — the answer had already been shown, so no credit this time. Next try is worth Answer shown after three tries — this one no longer earns credit. You chose to see the answer — this one earns no credit.
Why
Multiplying by the conjugate \dfrac{\sqrt x+3}{\sqrt x+3} turns the numerator into x-9, which cancels: \dfrac{1}{\sqrt x+3}. At x=16: \dfrac{1}{4+3}=\dfrac17\approx0.1429.
Correct answer
Also accepted:
Why
Multiply top and bottom by the conjugate sqrt(x)+3 first -- the numerator becomes a difference of squares, x-9, which cancels.
Which factorization correctly factors 2x^2+5x-3?
Right on try — credit. Right — the answer had already been shown, so no credit this time. Next try is worth Answer shown after three tries — this one no longer earns credit. You chose to see the answer — this one earns no credit.
Expanding this gives 2x^2-6x+x-3=2x^2-5x-3 -- the middle term's sign comes out wrong.
Expanding this gives 2x^2+3x-2x-3=2x^2+x-3 -- the middle term doesn't match.
Expanding this gives 2x^2+2x-3x-3=2x^2-x-3 -- the middle term's sign comes out wrong.
Why
(2x-1)(x+3)=2x^2+6x-x-3=2x^2+5x-3, matching every term of the original.
Expand each candidate using FOIL and check whether it actually reproduces 2x^2+5x-3 term by term.
(2x-1)(x+3)
A limit expression has a square root in the numerator and a plain polynomial (no square root) in the denominator, both equal to 0 at the point in question. Which technique is the right first move?
Right on try — credit. Right — the answer had already been shown, so no credit this time. Next try is worth Answer shown after three tries — this one no longer earns credit. You chose to see the answer — this one earns no credit.
A square root doesn't factor the way a polynomial does -- there's no pair of polynomial factors hiding inside a radical term.
A decimal approximation loses exactness and doesn't address why substitution fails -- it also doesn't cancel anything.
The conjugate trick handles exactly this shape, turning the radical into a difference of squares that cancels the trouble spot.
Why
A radical in the numerator calls for rationalizing (multiplying by its conjugate), not factoring -- factoring only applies to polynomial expressions, and there's no polynomial factor pattern to exploit here.
A radical term doesn't factor like a polynomial does -- the standard move for a lone square root is multiplying by its conjugate.
Rationalize the numerator by multiplying by its conjugate
Quick review
Practice from earlier coursework, pulled in for this topic.
From College Algebra
Factor 9x^2 - 12x + 4 as a perfect square. Type the result (using parentheses, * for multiplication, and ^ for exponents).
Right on try — credit. Right — the answer had already been shown, so no credit this time. Next try is worth Answer shown after three tries — this one no longer earns credit. You chose to see the answer — this one earns no credit.
Any equivalent form counts — 2x+3, 3 + 2·x, … Use the keypad below the field for fractions, roots, and other symbols.
Why
9x^2=(3x)^2 and 4=2^2 are both perfect squares, and 2(3x)(2)=12x matches the middle term. Since the middle term is negative, use the minus pattern: (3x-2)^2.
Correct answer
Also accepted:
Why
Check both ends are perfect squares first: 9x^2=(3x)^2, 4=2^2. Then confirm 2(3x)(2) matches the middle term's size.
From College Algebra
Factor 4x^2 - 4x - 15.
Right on try — credit. Right — the answer had already been shown, so no credit this time. Next try is worth Answer shown after three tries — this one no longer earns credit. You chose to see the answer — this one earns no credit.
Expanding gives 4x^2+4x-15 — the middle sign is flipped from the target.
Expanding gives 4x^2+7x-15, not matching the original middle term.
Expanding gives 4x^2-17x-15, not matching.
Why
Look for two numbers multiplying to ac=4\times(-15)=-60 and adding to b=-4: that's -10 and 6. Split: 4x^2-10x+6x-15=2x(2x-5)+3(2x-5)=(2x-5)(2x+3).
Use the ac-method: find two numbers multiplying to 4\times(-15) and adding to -4.
Quick review
Practice from earlier coursework, pulled in for this topic.
From College Algebra
Clear the radical from the denominator of \frac{3}{\sqrt{7} - \sqrt{2}}.
Right on try — credit. Right — the answer had already been shown, so no credit this time. Next try is worth Answer shown after three tries — this one no longer earns credit. You chose to see the answer — this one earns no credit.
You correctly built the conjugate product but forgot to keep the 3 from the original numerator.
The conjugate carries the opposite middle sign from the original denominator, not the same one.
Recheck the denominator after multiplying by the conjugate: (\sqrt7)^2-(\sqrt2)^2=7-2=5, not something that cancels the numerator's 3.
Why
Multiply top and bottom by the conjugate \sqrt7+\sqrt2: the denominator becomes 7-2=5, and the numerator becomes 3(\sqrt7+\sqrt2), giving \frac{3(\sqrt7+\sqrt2)}{5}.
Multiply top and bottom by the conjugate of the denominator, then simplify using difference of squares.