Solving Polynomial Equations
Every question with a fully worked solution.
- Solving Polynomial Equations - Video - Finding all the Zeros of a Polynomial Watch
Theory
To find the zeros of a polynomial, factorise it fully and set each factor equal to zero. This Year 10A maths page covers grouping, quadratics in \(x^2\), the factor theorem with division, the quadratic formula for leftover quadratics, quartics and problems in context.
A zero of \(P(x)\) is a value of \(x\) that makes \(P(x) = 0\). Zeros are the solutions (roots) of \(P(x) = 0\) and the \(x\)-intercepts of \(y = P(x)\).
The null factor law: if a product is zero, one of the factors is zero. So once \(P(x)\) is written as a product, each factor gives a zero.
A polynomial of degree \(n\) has at most \(n\) real zeros. It can have fewer: a factor such as \(x^2 + 1\) is never zero, and a repeated factor such as \((x - 2)^2\) gives one zero counted twice.
When \(P(x)\) does not factorise by inspection, the factor theorem finds a linear factor, division gives the quotient, and the process repeats until only quadratics or linear factors are left.
This prepares you for Mathematical Methods and Specialist Mathematics in Years 11 and 12.
| Pattern | Factorises as |
|---|---|
| \(x^3 - a^2x\) | \(x(x - a)(x + a)\) |
| \(x^3 + px^2 + qx + pq\) style | group: \((x + p)(x^2 + q)\) |
| \(x^4 + px^2 + q\) | quadratic in \(u = x^2\) |
| \(x^3 - a^3\) | \((x - a)(x^2 + ax + a^2)\) |
How to find the zeros of a polynomial
- Look for shortcuts: a common factor (often \(x\)), grouping in pairs, or a quadratic in \(x^2\).
- Otherwise test factors of the constant term until \(P(a) = 0\). With a leading coefficient other than \(1\), also try fractions.
- Divide by \((x - a)\). If two zeros are known, divide by their product in one step.
- Finish the quadratic quotient by factorising, completing the square or the quadratic formula. Note if it has no real zeros.
- List every zero, marking repeated ones, and check there are no more than the degree allows.
In context
- Build the equation from the situation, simplify it to \(P(x) = 0\) (divide through by a common number), then solve.
- Check each zero against the situation: lengths must be positive, so reject values that make any dimension zero or negative.
- Show that a polynomial has exactly \(k\) real zeros by factorising and showing any remaining quadratic is always positive.
(i) \(x(x + 4)(2x - 3)\)
(ii) \(x^3 + 3x^2 - 9x - 27\)
(iii) \(x^4 - 5x^2 - 36\)
(i) Already factorised: set each factor to zero.
| \(x = 0,\quad x + 4 = 0,\quad 2x - 3\) | \(=\) | \(0\) |
| \(x\) | \(=\) | \(\boldsymbol{0},\ \boldsymbol{-4},\ \boldsymbol{\dfrac{3}{2}}\) |
(ii) Group the terms in pairs.
| \(x^3 + 3x^2 - 9x - 27\) | \(=\) | \(x^2(x + 3) - 9(x + 3)\) |
| \(\) | \(=\) | \((x + 3)(x^2 - 9)\) |
| \(\) | \(=\) | \((x + 3)^2(x - 3)\) |
Zeros \(\boldsymbol{x = -3}\) (a double zero) and \(\boldsymbol{x = 3}\).
(iii) Only \(x^4\) and \(x^2\) appear: let \(u = x^2\).
| \(u^2 - 5u - 36\) | \(=\) | \(0\) |
| \((u - 9)(u + 4)\) | \(=\) | \(0\) |
| \(x^2 = 9 \text{ or } x^2\) | \(=\) | \(-4\) |
\(x^2 = -4\) has no real solution. So \(\boldsymbol{x = \pm 3}\).
Test factors of \(4\): \(P(1) = -1\), \(P(2) = 8 - 12 + 4 = 0\). So \((x - 2)\) is a factor.
| \(x^2\) | \(+2x\) | \(-2\) | ||
| \(x - 2\) | \(x^3\) | \(+0x^2\) | \(-6x\) | \(+4\) |
| \(-\) | \(x^3\) | \(-2x^2\) | ||
| \(2x^2\) | \(-6x\) | |||
| \(-\) | \(2x^2\) | \(-4x\) | ||
| \(-2x\) | \(+4\) | |||
| \(-\) | \(-2x\) | \(+4\) | ||
| \(0\) |
The quotient \(x^2 + 2x - 2\) does not factorise with whole numbers, so use the quadratic formula.
| \(x\) | \(=\) | \(\dfrac{-2 \pm \sqrt{4 + 8}}{2}\) |
| \(\) | \(=\) | \(\dfrac{-2 \pm 2\sqrt3}{2} = -1 \pm \sqrt3\) |
The zeros are \(\boldsymbol{x = 2}\) and \(\boldsymbol{x = -1 \pm \sqrt3}\).
Test the easy values: \(P(1) = 1 + 1 - 13 - 1 + 12 = 0\) and \(P(-1) = 1 - 1 - 13 + 1 + 12 = 0\).
So \((x - 1)(x + 1) = x^2 - 1\) is a factor. Divide by it in one step.
| \(x^2\) | \(+x\) | \(-12\) | |||
| \(x^2 - 1\) | \(x^4\) | \(+x^3\) | \(-13x^2\) | \(-x\) | \(+12\) |
| \(-\) | \(x^4\) | \(+0x^3\) | \(-x^2\) | ||
| \(x^3\) | \(-12x^2\) | \(-x\) | |||
| \(-\) | \(x^3\) | \(+0x^2\) | \(-x\) | ||
| \(-12x^2\) | \(+0x\) | \(+12\) | |||
| \(-\) | \(-12x^2\) | \(+0x\) | \(+12\) | ||
| \(0x\) | \(+0\) |
| \(P(x)\) | \(=\) | \((x^2 - 1)(x^2 + x - 12)\) |
| \(\) | \(=\) | \((x - 1)(x + 1)(x + 4)(x - 3)\) |
\(x = \boldsymbol{-4}, \boldsymbol{-1}, \boldsymbol{1}, \boldsymbol{3}\): four zeros, the most a quartic can have.
(i) Show that \(x^3 - 16x^2 + 60x - 45 = 0\).
(ii) Find all possible values of \(x\), correct to two decimal places where needed.
(i) The base is \((20 - 2x)\) by \((12 - 2x)\) and the height is \(x\).
| \(x(20 - 2x)(12 - 2x)\) | \(=\) | \(180\) |
| \(4x^3 - 64x^2 + 240x - 180\) | \(=\) | \(0\) |
| \(x^3 - 16x^2 + 60x - 45\) | \(=\) | \(0 \quad (\div 4)\) |
(ii) \(P(1) = 1 - 16 + 60 - 45 = 0\), so \((x - 1)\) is a factor: \(P(x) = (x - 1)(x^2 - 15x + 45)\).
| \(x\) | \(=\) | \(\dfrac{15 \pm \sqrt{225 - 180}}{2} = \dfrac{15 \pm 3\sqrt5}{2}\) |
| \(\) | \(\approx\) | \(4.15 \text{ or } 10.85\) |
The width \(12 - 2x\) must be positive, so \(0 \lt x \lt 6\). Reject \(10.85\). Both \(\boldsymbol{x = 1}\) cm and \(\boldsymbol{x \approx 4.15}\) cm give a box of volume \(180\) cm\(^3\).
Common pitfalls
Frequently asked questions
How do you find the zeros of a cubic polynomial?
Factorise it fully, using a common factor, grouping, or the factor theorem to find a linear factor and division to find the quadratic factor. Then set each factor equal to zero.
How many zeros can a polynomial have?
A polynomial of degree \(n\) has at most \(n\) real zeros. A cubic has \(1\), \(2\) or \(3\) distinct real zeros, and a quartic has at most \(4\).
What if the quadratic factor will not factorise?
Use the quadratic formula or complete the square. If \(b^2 - 4ac \lt 0\), that quadratic has no real zeros.
How do you solve x to the 4 minus 10x squared plus 9 equals 0?
Let \(u = x^2\) to get \(u^2 - 10u + 9 = 0\), so \(u = 1\) or \(u = 9\). Then \(x^2 = 1\) or \(9\), giving \(x = \pm1\) and \(x = \pm3\).
What is a double zero?
A zero that comes from a repeated factor, such as \(x = 2\) from \((x - 2)^2\). On the graph the curve touches the \(x\)-axis there instead of crossing it.