Dividing, factorising and finding unknowns

In General Maths you divided polynomials by long division and used two facts about dividing by $x - a$ (see polynomials↺):

• The remainder theorem: the remainder is $f(a)$.
• The factor theorem: if $f(a) = 0$, then $x - a$ is a factor.

Further Maths questions use the same two facts, but ask for more with them: all three factors of a cubic, two unknowns from a factor and a remainder, or a missing coefficient after dividing by a quadratic. This lesson starts with a faster way to divide, then takes each type in turn.

Adding and taking away polynomials

Add or take away like terms only: $x^3$ terms with $x^3$ terms, $x^2$ terms with $x^2$ terms, and so on. When a whole polynomial is multiplied by a number and then taken away, every one of its terms is multiplied, and every sign changes.

For example, with $P(x) = 4x^3 - 2x^2 + 5x - 1$ and $Q(x) = x^2 - 3x + 2$, to find $P(x) - 3Q(x)$:

• Multiply every term of $Q$ by 3: ${3Q(x) = 3x^2 - 9x + 6}$.
• Take away term by term: ${4x^3 + (-2 - 3)x^2 + (5 + 9)x + (-1 - 6)}$.
• Simplify: ${P(x) - 3Q(x) = 4x^3 - 5x^2 + 14x - 7}$.

Synthetic division

Long division works, but it is slow. Synthetic division does the same job when dividing by $x - a$, using only the coefficients:

1. Write the coefficients in order of power, with 0 for any missing power.
2. On the left, write $a$, the value that makes $x - a$ zero. For $x + 2$, that is $-2$.
3. Bring the first coefficient straight down.
4. Multiply it by $a$, write the product under the next coefficient, and add. Repeat to the end.
5. The last number is the remainder. The others are the coefficients of the quotient, which is one power lower than the polynomial.

Step through it:

The remainder always agrees with the remainder theorem. In the figure, $f(2) = 8 + 8 - 10 + 1 = 7$. So synthetic division gives you $f(a)$ and the quotient in one go.

Factorising a cubic completely

To factorise a cubic, you need one factor to start with. The question may give it to you, as "$f(-1) = 0$". If not, try the factors of the constant term in $f(a)$ until one gives 0. Then:

1. divide that factor out (synthetic division is quickest), leaving a quadratic;
2. factorise the quadratic (see factorising↺);
3. write $f(x)$ as the product of all three brackets.

A factor and a remainder together

Each fact you are given turns into one equation:

• ”$x - a$ is a factor” gives $f(a) = 0$;
• “the remainder on dividing by $x - a$ is $R$” gives $f(a) = R$.

Two unknowns need two facts. Solve the two equations simultaneously.

Dividing by a quadratic: compare coefficients

For any division, the polynomial equals the divisor times the quotient, plus the remainder. This is true for every value of $x$:

So when a question gives you the divisor, the quotient and the remainder, multiply out the right-hand side. Then match the coefficients of each power of $x$ with those of $f(x)$.

A quadratic from three of its values

$f(x) = px^2 + qx + r$ has three unknowns, so it takes three facts to find them. Each value such as $f(2) = 7$ gives one equation. Take one equation from another to get rid of a letter, just as with two unknowns.

For example, if $f(1) = 2$, $f(-1) = 6$ and $f(2) = 3$ for $f(x) = ax^2 + bx + c$:

• $f(1) = 2$ gives ${a + b + c = 2}$.
• $f(-1) = 6$ gives ${a - b + c = 6}$.
• $f(2) = 3$ gives ${4a + 2b + c = 3}$.
• Take the second from the first: ${2b = -4}$, so ${b = -2}$.
• Put $b = -2$ in the first: ${a + c = 4}$.
• Put $b = -2$ in the third: ${4a + c = 7}$.
• Take ${a + c = 4}$ from ${4a + c = 7}$: ${3a = 3}$, so ${a = 1}$.
• Then ${c = 4 - 1 = 3}$, and ${f(x) = x^2 - 2x + 3}$.

Your turn