Factorising Completely

If $ab = 0$ for two real numbers, what can you conclude about $a$ and $b$? Now if $ab = 12$, what can you conclude? The difference between those two answers is the entire reason we factorise before solving.

Solving a polynomial equation has two stages, and the second one is trivial. Factorise completely, then apply the null factor law: a product of real numbers is zero exactly when at least one of the factors is zero.

If two real numbers multiply to give zero, at least one of them must be zero. There is no other way to reach zero by multiplying.

$$AB = 0 \quad \iff \quad A = 0 \ \text{ or } \ B = 0$$

The same holds for any number of factors. So from

$$(x-1)(x+2)(2x-3) = 0$$

you get three separate linear equations, each solved on its own: $x = 1$, $x = -2$, $x = \tfrac{3}{2}$.

The condition that makes this work is that the right-hand side is zero. Nothing similar is true for any other number. From $(x-1)(x+2) = 4$ you cannot conclude that $x - 1 = 4$ or $x + 2 = 4$, because $4$ can be reached by multiplying in infinitely many ways, such as $2 \times 2$ or $8 \times \tfrac{1}{2}$.

Factorising completely means continuing until no factor can be broken down further. Two habits make it reliable.

Take out any common factor first. For $2x^3 - 8x$, pulling out $2x$ gives $2x(x^2 - 4) = 2x(x-2)(x+2)$. Testing candidates on the original would have worked too, but with larger numbers and more of them.

Know when to stop. Over the real numbers, a quadratic factor $ax^2 + bx + c$ is irreducible when its discriminant is negative:

$$b^2 - 4ac < 0 \quad \Longrightarrow \quad \text{no real zeroes, so no further real factors}$$

So $x^3 - 1 = (x-1)(x^2 + x + 1)$ is completely factorised: the quadratic has discriminant $1 - 4 = -3 < 0$. Attempting to push further is not merely unnecessary, it is impossible, and an answer that leaves the bracket intact is correct.

Check the discriminant before you go looking for more factors. It converts an open-ended search into a decided question.

The routine from Lesson 7, stated as a procedure:

Take out any common factor. Build the candidate list from the constant term. Test candidates until $P(a) = 0$. Write down the factor $(x-a)$. Divide to get the quotient. Factorise the quotient, which is a quadratic. Stop when nothing more factorises.

Solve $x^3 - 4x^2 + x + 6 = 0$.

The constant term is $6$, so test $\pm 1, \pm 2, \pm 3, \pm 6$. $P(-1) = -1 - 4 - 1 + 6 = 0$, so $x + 1$ is a factor. Dividing gives

$$x^3 - 4x^2 + x + 6 = (x+1)(x^2 - 5x + 6) = (x+1)(x-2)(x-3)$$

Applying the null factor law:

$$x = -1, \quad x = 2, \quad x = 3$$

Three solutions from a cubic, which is the most a cubic can have. Whether every cubic manages three is the subject of the next lesson.

A quartic needs the factor theorem twice, unless it has special structure.

The general route. Find one zero, divide to get a cubic, find a zero of that, divide to get a quadratic, factorise. Each division drops the degree by one.

The shortcut, when it applies. If only even powers appear, the quartic is a quadratic in disguise. Solve $x^4 - 5x^2 + 4 = 0$ by putting $u = x^2$:

$$u^2 - 5u + 4 = 0 \quad \Longrightarrow \quad (u-1)(u-4) = 0 \quad \Longrightarrow \quad u = 1 \ \text{ or } \ u = 4$$

Then convert back. $x^2 = 1$ gives $x = \pm 1$, and $x^2 = 4$ gives $x = \pm 2$. Four solutions: $-2, -1, 1, 2$.

The step people forget is the conversion back. $u$ was never what the question asked for, and each positive value of $u$ produces two values of $x$. A negative value of $u$ produces none, since no real number squares to a negative, which is how a quartic can end up with only two real solutions.

A factor can appear more than once. Factorising $x^3 - 3x^2 + 4$ starts with $P(-1) = -1 - 3 + 4 = 0$, so $x + 1$ is a factor, and dividing gives

$$x^3 - 3x^2 + 4 = (x+1)(x^2 - 4x + 4) = (x+1)(x-2)^2$$

The equation $P(x) = 0$ then has solutions $x = -1$ and $x = 2$. That is two solutions, not three: the repeated bracket $(x-2)^2$ is zero exactly when $x = 2$, and saying it twice does not produce a second number.

The repetition is still real information, and it is recorded by saying that $2$ is a zero of multiplicity two, or a double root. Lesson 11 shows that multiplicity is visible on the graph: the curve touches the horizontal axis at a double root instead of crossing it.

When testing candidates, a zero you have already used is worth testing again on the quotient. If $P(a) = 0$ and the quotient also vanishes at $a$, the factor is repeated.

Five equations. Work each one, then reveal the answer.

1. 1 Solve $(x-4)(2x+6) = 0$.

   Show answer
   Set each bracket to zero and solve separately.$x = 4$ or $x = -3$

2. 2 Solve $x^3 - 9x = 0$.

   Show answer
   Common factor first: $x(x-3)(x+3) = 0$.$x = 0, 3, -3$

3. 3 Solve $(x+5)^2 = 0$. How many solutions?

   Show answer
   A repeated factor is zero at one value only.$x = -5$, one solution

4. 4 Is $x^2 + x + 5$ factorisable over the reals?

   Show answer
   Discriminant $1 - 20$.No, $\Delta = -19 < 0$

5. 5 Solve $x^4 - 16 = 0$ over the real numbers.

   Show answer
   $(x^2-4)(x^2+4) = 0$; the second bracket has no real zeroes.$x = \pm 2$