Factoring by Substitution

1 MIN READ · PRACTISE AT YOUR OWN PACE · FREE

Idea

Sometimes an expression is quadratic in a repeated object even though it is not quadratic in $x$ . Name that object with a temporary variable, factor, and then substitute back. This reduces visual complexity without changing the algebra.

Visual guide

VISUAL GUIDEA quadratic in x²

x⁴ − 5x² + 4

For x⁴ − 5x² + 4, the substitution u = x² gives zeros u = 1 and 4. Each positive u produces two x values, so the original graph crosses at −2, −1, 1 and 2.

Method

For $x^4+bx^2+c$ , put $u=x^2$ . The expression becomes $u^2+bu+c$ . When solving over the reals, remember that $u=x^2\ge0$ .

Worked example

Factor $x^4-5x^2+4$ . With $u=x^2$ ,

u^2-5u+4=(u-1)(u-4).

Substitute back to get $(x^2-1)(x^2-4)$ , then factor both differences of squares:

(x-1)(x+1)(x-2)(x+2).

PAUSE & THINKA quick check, not a grade

Try it yourself.

Choose an answer and explain your reasoning to yourself. Use a hint if you get stuck.

Which substitution makes

x^4-5x^2+4

quadratic?

u = x²

u = x⁴

u = 1/x

Hint 1 · Find a starting point

The exponents are 4, 2 and 0.

Hint 2 · Take the next step

Look for one power whose square produces x⁴.

Show the reasoning

Answer: u = x²

With u=x², the expression becomes u²−5u+4 = (u−1)(u−4).

Before moving on: what would make one of the other answers wrong? Saying why is part of understanding.

Common mistake

Do not stop a solution at $u=1,4$ when the question asks for $x$ . Each positive square value gives two real roots.

Check your understanding

Factor $x^4+3x^2+2$ over the reals.

Show answer

$(x^2+1)(x^2+2)$ . Neither quadratic has real zeros.

MAKE IT YOURS

Pause before the next idea.

Can you explain how the visualization connects to this lesson’s goal? If a step still feels uncertain, put this lesson on your review list and try the check again another day.

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