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Factoring Quadratics for the SAT

Factoring is the fastest route through most SAT quadratics, and it is also how the test hides the answer in plain sight. Once an expression is written as a product of two binomials, the solutions are visible without any formula. The skill is finding the right pair of numbers fast.

The rule

Find two numbers that multiply to the constant and add to the middle coefficient, then remember that a factor of x minus 4 gives the solution positive 4.

For a quadratic that starts with x squared, you need two numbers that multiply to the constant term and add to the middle coefficient. Take x squared plus 7x plus 12. Pairs that multiply to 12 are 1 and 12, 2 and 6, and 3 and 4. The pair that adds to 7 is 3 and 4, so the expression factors as the quantity x plus 3 times the quantity x plus 4.

Signs follow from the signs of the constant and the middle term. If the constant is positive, both numbers share the sign of the middle coefficient. So x squared minus 7x plus 12 factors as the quantity x minus 3 times the quantity x minus 4. If the constant is negative, the two numbers have opposite signs, and the larger absolute value carries the sign of the middle term. So x squared minus x minus 12 factors as the quantity x minus 4 times the quantity x plus 3.

Once factored, use the zero product rule: if two things multiply to zero, at least one of them is zero. From the quantity x minus 4 times the quantity x plus 3 equals zero, you get x equals 4 or x equals negative 3. Notice the sign flip. A factor of x minus 4 corresponds to a solution of positive 4, which is the single most common place students report the wrong answer.

When the leading coefficient is not 1, first check for a common factor. In 2x squared plus 10x plus 12, pull out the 2 to get 2 times the quantity x squared plus 5x plus 6, then factor the inside normally into the quantity x plus 2 times the quantity x plus 3. If nothing factors out, as in 3x squared plus 8x plus 4, find two numbers that multiply to a times c, which is 12, and add to 8. Those are 6 and 2, so split the middle term: 3x squared plus 6x plus 2x plus 4, group into 3x times the quantity x plus 2 plus 2 times the quantity x plus 2, and factor to the quantity 3x plus 2 times the quantity x plus 2.

If no factor pair appears within about twenty seconds, stop looking and use the quadratic formula or the calculator. Many SAT quadratics are deliberately not factorable over the integers, and the test is partly measuring whether you know when to switch tools instead of grinding.

The trap

Reading the solutions straight off the factors without flipping the signs, so the quantity x plus 3 gets reported as a solution of 3.

Try it

Try one
Real question
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The graph of y = f(x) is shown in the xy-plane. For which values of x is f(x) = 2?

What is the solution set of √x = x - 2?

  • A) No real solution
  • B) x = 1 and x = 4
  • C) x = 4 ✓
  • D) x = 1

Why: Square both sides: x = (x - 2)² = x² - 4x + 4, so x² - 5x + 4 = 0 and (x - 1)(x - 4) = 0. That gives candidates x = 1 and x = 4, and squaring can create roots that do not satisfy the original equation, so check both. For x = 4, √4 = 2 and 4 - 2 = 2, so it works; for x = 1, √1 = 1 but 1 - 2 = -1, so it fails. The answer x = 1 and x = 4 reports both candidates without the check. The answer x = 1 keeps the extraneous root and drops the valid one. The answer No real solution rejects x = 4, which does satisfy the equation.

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Questions students ask

What do I do if a quadratic will not factor?

Use the quadratic formula, which works on every quadratic. Compute the discriminant first, because a negative value means there are no real solutions and you can stop. Completing the square is a third option and is the one to use when the question asks for the vertex.

How do I factor when the leading coefficient is not 1?

First try pulling out a common factor from all three terms. If that fails, find two numbers that multiply to the first coefficient times the constant and add to the middle coefficient, split the middle term using them, and factor by grouping in pairs.

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