From Zero to Engineer

Horner's method and the remainder theorem

Writing a polynomial in nested form, evaluating it without computing any powers, and the remainder theorem: the remainder on division by x − a equals f(a).

Evaluating a polynomial directly means raising the argument to successive powers, and every one of those multiplications is a chance to slip. The nested form rearranges the computation so that only multiplications by x and additions are left - it is faster and safer, on a calculator and in a program alike. The second half of the lesson shows what it is good for: since the remainder on dividing f(x) by x − a equals f(a), a single evaluation replaces a whole long division.

Nested form of a polynomial

We rewrite the polynomial so that each step is a multiplication by x followed by adding the next coefficient. It is built up starting from the highest power.

f(x) = 5x^{3} + 2x^{2} - 3x + 6

1^{\circ}\quad 5x + 2

4^{\circ}\quad [(5x + 2)x - 3]x

2^{\circ}\quad (5x + 2)x

5^{\circ}\quad [(5x + 2)x - 3]x + 6

3^{\circ}\quad (5x + 2)x - 3

f(x) = [(5x + 2)x - 3]x + 6

f(4) = [(5\cdot 4 + 2)\cdot 4 - 3]\cdot 4 + 6 = [22\cdot 4 - 3]\cdot 4 + 6 = 85\cdot 4 + 6 = 340 + 6 = 346

f(2) = [(5\cdot 2 + 2)\cdot 2 - 3]\cdot 2 + 6 = [12\cdot 2 - 3]\cdot 2 + 6 = 21\cdot 2 + 6 = 42 + 6 = 48

f(-1) = [(5\cdot(-1) + 2)\cdot(-1) - 3]\cdot(-1) + 6 = [3 - 3]\cdot(-1) + 6 = 6

When a power is missing, write it in with a zero coefficient - otherwise the nesting comes out wrong.

f(x) = 3x^{4} + 2x^{2} - 4x + 5 = 3x^{4} + 0x^{3} + 2x^{2} - 4x + 5

1^{\circ}\quad 3x + 0

5^{\circ}\quad [(3x + 0)x + 2]x - 4

2^{\circ}\quad (3x + 0)x

6^{\circ}\quad \{[(3x + 0)x + 2]x - 4\}x

3^{\circ}\quad (3x + 0)x + 2

7^{\circ}\quad \{[(3x + 0)x + 2]x - 4\}x + 5

4^{\circ}\quad [(3x + 0)x + 2]x

f(x) = \{[(3x + 0)x + 2]x - 4\}x + 5

f(2) = \{[(3\cdot 2 + 0)\cdot 2 + 2]\cdot 2 - 4\}\cdot 2 + 5 = \{14\cdot 2 - 4\}\cdot 2 + 5 = 24\cdot 2 + 5 = 53

Write the polynomial in nested form and evaluate it for the given value of x.

\text{a)}\quad f(x) = 4x^{3} + 3x^{2} + 2x - 4

x = 2

f(x) = [(4x + 3)x + 2]x - 4

f(2) = [(4\cdot 2 + 3)\cdot 2 + 2]\cdot 2 - 4 = [22 + 2]\cdot 2 - 4 = 48 - 4 = 44

4\cdot 2^{3} + 3\cdot 2^{2} + 2\cdot 2 - 4 = 44

\text{b)}\quad f(x) = 2x^{4} + x^{3} - 3x^{2} + 5x - 6

x = 3

f(x) = \{[(2x + 1)x - 3]x + 5\}x - 6

\begin{aligned}f(3) &= \{[(2\cdot 3 + 1)\cdot 3 - 3]\cdot 3 + 5\}\cdot 3 - 6 = \{[21 - 3]\cdot 3 + 5\}\cdot 3 - 6 = \\&= \{54 + 5\}\cdot 3 - 6 = 177 - 6 = 171\end{aligned}

2\cdot 3^{4} + 3^{3} - 3\cdot 3^{2} + 5\cdot 3 - 6 = 171

\text{c)}\quad f(x) = x^{4} - 3x^{3} + 2x - 3

x = 5

f(x) = x^{4} - 3x^{3} + 0x^{2} + 2x - 3 = \{[(x - 3)x + 0]x + 2\}x - 3

f(5) = \{[(5 - 3)\cdot 5 + 0]\cdot 5 + 2\}\cdot 5 - 3 = 52\cdot 5 - 3 = 260 - 3 = 257

5^{4} - 3\cdot 5^{3} + 2\cdot 5 - 3 = 257

\text{d)}\quad f(x) = 2x^{4} - 5x^{3} - 3x^{2} + 4

x = 4

f(x) = 2x^{4} - 5x^{3} - 3x^{2} + 0x + 4 = \{[(2x - 5)x - 3]x + 0\}x + 4

f(4) = \{[(2\cdot 4 - 5)\cdot 4 - 3]\cdot 4 + 0\}\cdot 4 + 4 = 36\cdot 4 + 4 = 148

2\cdot 4^{4} - 5\cdot 4^{3} - 3\cdot 4^{2} + 4 = 148

The remainder theorem

\begin{aligned}\frac{f(x)}{x - a} &= q(x) + \frac{R}{x - a} \;\Big/ \cdot (x - a) \\f(x) &= (x - a)\cdot q(x) + R\end{aligned}

f(a) = 0\cdot q(a) + R \quad\Rightarrow\quad f(a) = R

If the polynomial f(x) is divided by x − a, the remainder is f(a).

Determine the remainder.

(x^{3} + 3x^{2} - 13x - 10) : (x - 3)

[(x + 3)x - 13]x - 10

f(3) = [(3 + 3)\cdot 3 - 13]\cdot 3 - 10 = 15 - 10 = 5

Long division gives the same result, only the long way round.

(x^{3} + 3x^{2} - 13x - 10) : (x - 3) = x^{2} + 6x + 5\ \text{r.}\ 5

(x^{3} + 3x^{2} - 13x - 10) = (x - 3)(x^{2} + 6x + 5) + 5

Determine the remainder in each of the following cases.

\text{a)}\quad (5x^{3} - 4x^{2} - 3x + 6) : (x - 2)

[(5x - 4)x - 3]x + 6

f(2) = [(5\cdot 2 - 4)\cdot 2 - 3]\cdot 2 + 6 = 18 + 6 = 24

\text{b)}\quad (4x^{3} - 3x^{2} + 5x - 3) : (x - 4)

[(4x - 3)x + 5]x - 3

f(4) = [(4\cdot 4 - 3)\cdot 4 + 5]\cdot 4 - 3 = 57\cdot 4 - 3 = 228 - 3 = 225

\text{c)}\quad (x^{3} - 2x^{2} - 3x + 5) : (x - 5)

[(x - 2)x - 3]x + 5

f(5) = [(5 - 2)\cdot 5 - 3]\cdot 5 + 5 = 60 + 5 = 65

\text{d)}\quad (2x^{3} + 3x^{2} - x + 4) : (x + 2)

[(2x + 3)x - 1]x + 4

f(-2) = [(2\cdot(-2) + 3)\cdot(-2) - 1]\cdot(-2) + 4 = -2 + 4 = 2

\text{e)}\quad (3x^{3} - 11x^{2} + 10x - 12) : (x - 3)

[(3x - 11)x + 10]x - 12

f(3) = [(3\cdot 3 - 11)\cdot 3 + 10]\cdot 3 - 12 = 12 - 12 = 0

A remainder of zero means that x − 3 divides this polynomial exactly.

Frequently asked questions

What is Horner's method?

On writing the polynomial in nested form, e.g. 5x³ + 2x² − 3x + 6 = [(5x + 2)x − 3]x + 6. Only multiplications by x and additions are left, with no powers to compute.

What if a power is missing from the polynomial?

Write it in with a coefficient of zero. In 3x⁴ + 2x² − 4x + 5 the x³ term is missing, so write 3x⁴ + 0x³ + 2x² − 4x + 5 and nest that. Skip this step and the nested form comes out wrong.

What does the remainder theorem say?

That the remainder on dividing f(x) by x − a equals f(a). It follows from f(x) = (x − a)·q(x) + R: substituting x = a kills the first term and leaves f(a) = R.

How do you check whether x − a divides a polynomial exactly?

Evaluate f(a). If it comes out zero, the division is exact. For 3x³ − 11x² + 10x − 12 and the binomial x − 3 we have f(3) = 0, so x − 3 is a factor.

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Rearranging formulae and evaluating functions

The factor theorem and factoring polynomials