Solve polynomial equations using factoring where applicable. | Nonlinear equations in one variable | SAT Math

Higher-degree polynomials usually solve by factoring and setting each factor to zero.

核心知识

For polynomial equations of degree 3 or higher, the main strategy is to factor: pull out a greatest common factor, then use grouping or pattern recognition (difference of squares, sum/difference of cubes) to break the polynomial into linear and quadratic factors. Set each factor to zero.

深入理解

The SAT rarely gives you a cubic or quartic that can't be factored. When you see a higher-degree polynomial equation, your first move should be to look for a common factor.

If every term contains $𝑥$ , factor out the highest power of $𝑥$ that's common. This immediately gives you $𝑥 = 0$ as one solution and reduces the degree of what remains.

For four-term polynomials, try factoring by grouping: pair the first two and last two terms, factor each pair, then factor out the common binomial.

Recognize special patterns:

$𝑎 2 - 𝑏 2 = (𝑎 - 𝑏) (𝑎 + 𝑏)$
$𝑎 3 + 𝑏 3 = (𝑎 + 𝑏) (𝑎 2 - 𝑎 𝑏 + 𝑏 2)$
$𝑎 3 - 𝑏 3 = (𝑎 - 𝑏) (𝑎 2 + 𝑎 𝑏 + 𝑏 2)$

Some problems are "quadratic in disguise": $𝑥 4 - 5 𝑥 2 + 4 = 0$ becomes $𝑢 2 - 5 𝑢 + 4 = 0$ where $𝑢 = 𝑥 2$ .

分步讲解

Set the equation equal to zero.
Factor out the greatest common factor (GCF) from all terms.
Factor the remaining polynomial using grouping, special patterns, or substitution.
Set each factor equal to zero and solve.
Collect all solutions.

常见误解

Dividing both sides by $𝑥$ instead of factoring out $𝑥$ . Dividing by $𝑥$ loses the solution $𝑥 = 0$ .
Stopping after finding one factor. A cubic can have up to three real solutions.
Not recognizing a "quadratic in disguise" — for instance, treating $𝑥 4 - 5 𝑥 2 + 4 = 0$ as unsolvable when it factors as $(𝑥 2 - 1) (𝑥 2 - 4) = 0$ .

题目

示例解析

What are all solutions to $𝑥 3 - 4 𝑥 = 0$ ?

选择一个答案查看解析