SQA Higher Mathematics

Polynomials

A polynomial in $x$ is an expression made of powers of $x$ with whole-number indices, each multiplied by a constant called a coefficient, such as $4x^3 - x^2 + 7x - 5$. The highest power is its degree, so this one is a cubic, and the coefficient of that power is the leading coefficient. A polynomial of degree 4 is a quartic, and the constant term is the term with no $x$.

A number $x = a$ for which $f(a) = 0$ is a root of $f(x) = 0$ and a zero of the function $f$. Finding roots is the same problem as finding linear factors, which is what the factor theorem says.

The remainder theorem

When a polynomial $f(x)$ is divided by $(x - h)$, the remainder is $f(h)$. The reason is that dividing gives $f(x) = (x - h)q(x) + r$ with $r$ a constant, and putting $x = h$ makes the first term vanish, leaving $f(h) = r$.

So the remainder can be found without any division. To find the remainder when $f(x) = x^3 + 4x^2 - 2x + 5$ is divided by $(x + 2)$, put $x = -2$:

For a divisor such as $(2x - 1)$, the value to substitute is the one that makes the divisor zero, here $x = \frac{1}{2}$.

The factor theorem

$(x - h)$ is a factor of $f(x)$ exactly when $f(h) = 0$. This is the remainder theorem with remainder zero, and it works both ways: a zero of the remainder means a factor, and a factor means a root. In the same way $(ax - b)$ is a factor when $f\!\left(\frac{b}{a}\right) = 0$.

The theorem is also used to find unknown coefficients. If $(x + 3)$ is a factor of $x^3 + ax^2 - 4x + 6$, then the polynomial must be zero at $x = -3$:

If the remainder is given instead, set $f(h)$ equal to it. When $f(x) = 2x^3 + x^2 - 13x + k$ leaves remainder 4 on division by $(x - 2)$, then $f(2) = 16 + 4 - 26 + k = k - 6 = 4$, so $k = 10$.

Finding a first root

To factorise a cubic or quartic there has to be a first factor, and it is found by trial. If the coefficients are whole numbers, any whole-number root divides the constant term, so these are the values to try first: for a constant term of 6, the values $\pm 1, \pm 2, \pm 3, \pm 6$. When the leading coefficient is not 1, a fractional root $\frac{p}{q}$ can occur, where $p$ divides the constant term and $q$ divides the leading coefficient. Try the simplest values first, usually $x = 1$ and $x = -1$.

Synthetic division

Once one root $h$ is known, the remaining factor is found by dividing by $(x - h)$. Synthetic division does this using only the coefficients, written in order with zeros for any missing power.

To divide $x^3 - 2x^2 - 5x + 6$ by $(x - 1)$, use $h = 1$ and the coefficients $1, -2, -5, 6$. The first coefficient is brought down. Each result is then multiplied by $h$ and added to the next coefficient.

The last entry is the remainder, and it is 0, which confirms that $(x - 1)$ is a factor. The other entries are the coefficients of the quotient, starting one degree lower: $x^2 - x - 6$. Therefore

The same table gives the remainder when it is not zero, which is why it can be used for the remainder theorem too. When the divisor is $(x + 2)$, use $h = -2$, and the multiplier in the table is $-2$.

Factorising a cubic with a leading coefficient other than 1

For $2x^3 - 3x^2 - 11x + 6$, try $x = 3$: $54 - 27 - 33 + 6 = 0$, so $(x - 3)$ is a factor. Dividing synthetically with $h = 3$ and coefficients $2, -3, -11, 6$:

The quotient is $2x^2 + 3x - 2$, which factorises as $(2x - 1)(x + 2)$. So

Factorising a quartic

A quartic needs two roots found by trial, or one root and then a cubic. Take $f(x) = x^4 - 3x^3 - 7x^2 + 15x + 18$. Then $f(-1) = 1 + 3 - 7 - 15 + 18 = 0$, so $(x + 1)$ is a factor. Dividing by it with $h = -1$:

The quotient is the cubic $x^3 - 4x^2 - 3x + 18$. Trying $x = 3$ gives $27 - 36 - 9 + 18 = 0$, so divide again with $h = 3$:

This leaves $x^2 - x - 6 = (x - 3)(x + 2)$. Altogether

A check is to compare the constant term. Multiplying the constants of the factors gives $1 \times 2 \times 9 = 18$, which matches. The repeated factor $(x - 3)^2$ means $x = 3$ is a repeated root, and the graph touches the $x$-axis there instead of crossing it.

Factorising an expression that is not fully given

Sometimes the expression is given with a factor already known, and the job is to find the rest. If $(x - 2)$ is a factor of $x^3 + 2x^2 - 5x - 6$, check by $f(2) = 8 + 8 - 10 - 6 = 0$, then divide:

The quotient $x^2 + 4x + 3 = (x + 1)(x + 3)$, so the full factorisation is $(x - 2)(x + 1)(x + 3)$.

If the polynomial does not factorise into linear factors over the integers, the quadratic left after the first division may not factorise. Then the discriminant decides whether it has real roots, and the quadratic formula gives them when it does.