Reducing a harder equation to a simpler one

The equation $x^4-5x^2+4=0$ has degree four, but one structure repeats: $x^4$ is the square of $x^2$ . Writing $t=x^2$ produces a familiar quadratic. We must then return from $t$ to every possible $x$ , rather than answer a question about a different variable.

We need factoring and quadratic equations. We work over the real numbers; complex solutions are outside this lesson.

A substitution comes with a restriction

A biquadratic equation has the form $ax^4+bx^2+c=0$ , $a\ne0$ . Substituting $t=x^2$ gives $at^2+bt+c=0$ , necessarily with $t\ge0$ .

Here $t^2-5t+4=0$ , or $(t-1)(t-4)=0$ . Thus $t=1$ or 4. This is not yet the final answer:

x^2=1\quad\text{or}\quad x^2=4.

The first gives $x=\pm1$ , the second $x=\pm2$ . All four fit. Check: when $x^2=1$ , the original expression is $1-5+4=0$ ; when $x^2=4$ , it is $16-20+4=0$ .

The substitution t=x² requires t ≥ 0. Both t values are allowed. Returning gives all four x values: −2, −1, 1, 2; the negative sign must not be omitted.

The substitution is not one-to-one: 1 and −1 give the same $t=1$ . That is why the return journey branches.

A root for t may give no real x

For $x^4+3x^2-4=0$ , the same substitution gives $(t-1)(t+4)=0$ . The number $t=-4$ correctly solves the quadratic in $t$ , but cannot be the square of a real $x$ . Reject it because $t\ge0$ ; only $x=\pm1$ remain.

We can also substitute an entire repeated expression, such as $t=(x+1)^2$ . The restriction $t\ge0$ remains, but we must return to $(x+1)^2=t$ , not $x^2=t$ . Equations in quadratic form, OpenStax §2.6.

Factoring offers another way to reduce degree

Let $P(x)=x^3-4x^2+x+6$ . Trying $x=2$ gives $8-16+2+6=0$ , suggesting the factor $x-2$ . To find the rest, divide the polynomial by this binomial, successively removing its leading term.

First take $x^2$ : subtracting $(x-2)x^2=x^3-2x^2$ leaves $-2x^2+x+6$ . Next take $-2x$ : subtract $-2x^2+4x$ , leaving $-3x+6$ . The last quotient term, −3, removes the rest. Hence:

P(x)=(x-2)(x^2-2x-3).

The quotient is $x^2-2x-3$ ; the remainder is 0. Multiply back to check all four coefficients of $P$ . In general, division by a nonzero polynomial $B$ gives an identity $P=BQ+R$ , with $R=0$ or the degree of $R$ below that of $B$ . Polynomial division, OpenStax §5.4.

The remainder theorem explains substitution

For divisor $x-r$ , the remainder is constant: $P(x)=(x-r)Q(x)+R$ . Substitute $x=r$ ; the product vanishes, so $R=P(r)$ . This is the polynomial remainder theorem, also called Bézout’s theorem on remainders. Thus $r$ is a root if and only if $x-r$ divides the polynomial exactly. Remainder and factor theorems, OpenStax §5.5.

Here $x^2-2x-3=(x-3)(x+1)$ , so $P(x)=0$ gives $x=2,3,-1$ . We did not divide the equation by $x-2$ and lose 2: we considered the zero of every factor. Success with the trial number 2 does not imply that trial will always find a root quickly.

Branch · A checkDivision works even without a known root.

When the remainder is nonzero

For $P(x)=x^2+1$ , we have $P(x)=(x-2)(x+2)+5$ . The remainder 5 equals $P(2)$ , so 2 is not a root. The identity holds even at $x=2$ , although the numerical fraction $P(x)/(x-2)$ is undefined there. Polynomial identities and quotient domains are different questions.

Degree reduction works when we preserve the link to the original problem: a substitution’s restriction and every return branch, or every factor of a zero product. A shorter intermediate expression is useful together with that link.