Splitting a quartic into four linear factors: two roots found by trial, division by the product of the corresponding binomials, and the quadratic formula to finish.

A quartic is factored the same way as a cubic, only the step is done twice. You can divide twice by single binomials or, once two roots are known, divide straight away by their product - the second route is shorter, because one division drops the degree from four to two. Either way a quadratic is what remains, and that is a familiar computation.

Two divisions in a row

f(x) = 2x^{4} - x^{3} - 8x^{2} + x + 6 = \{[(2x - 1)x - 8]x + 1\}x + 6

x = 1 \;\Rightarrow\; \{[(2\cdot 1 - 1)\cdot 1 - 8]\cdot 1 + 1\}\cdot 1 + 6 = 0

(2x^{4} - x^{3} - 8x^{2} + x + 6) : (x - 1) = 2x^{3} + x^{2} - 7x - 6

$2x^{4}$	$- 2x^{3}$
0	$x^{3}$	$- 8x^{2}$
	$x^{3}$	$- x^{2}$
	0	$- 7x^{2}$	$+ x$
		$- 7x^{2}$	$+ 7x$
		0	$- 6x$	$+ 6$
			$- 6x$	$+ 6$
				$0$

(x - 1)(2x^{3} + x^{2} - 7x - 6) = 2x^{4} - x^{3} - 8x^{2} + x + 6

2x^{4} - x^{3} - 8x^{2} + x + 6 = (x - 1)(2x^{3} + x^{2} - 7x - 6)

the second bracket is g(x)

g(x) = 2x^{3} + x^{2} - 7x - 6 = [(2x + 1)x - 7]x - 6

x = -1 \;\Rightarrow\; [(2\cdot(-1) + 1)\cdot(-1) - 7]\cdot(-1) - 6 = 0

(2x^{3} + x^{2} - 7x - 6) : (x + 1) = 2x^{2} - x - 6

$2x^{3}$	$+ 2x^{2}$
0	$- x^{2}$	$- 7x$
	$- x^{2}$	$- x$
	0	$- 6x$	$- 6$
		$- 6x$	$- 6$
			$0$

(x + 1)(2x^{2} - x - 6) = 2x^{3} + x^{2} - 7x - 6

2x^{4} - x^{3} - 8x^{2} + x + 6 = (x - 1)(x + 1)(2x^{2} - x - 6)

x = \frac{1 \pm \sqrt{(-1)^{2} - 4\cdot 2\cdot(-6)}}{2\cdot 2} = \frac{1 \pm \sqrt{1 + 48}}{4} = \frac{1 \pm 7}{4}

x_{1} = 2

x_{2} = -\tfrac{3}{2}

2x^{2} - x - 6 = 2(x - 2)\left(x + \tfrac{3}{2}\right) = (x - 2)(2x + 3)

2x^{4} - x^{3} - 8x^{2} + x + 6 = (x - 1)(x + 1)(x - 2)(2x + 3)

One division by the product of the binomials

When two roots turn up in the trials, there is no need to divide twice. Divide by the product of the two binomials instead.

f(x) = 2x^{4} - 5x^{3} - 15x^{2} + 10x + 8 = \{[(2x - 5)x - 15]x + 10\}x + 8

x = 1 \;\Rightarrow\; \{[(2\cdot 1 - 5)\cdot 1 - 15]\cdot 1 + 10\}\cdot 1 + 8 = 0

x = -2 \;\Rightarrow\; \{[(2\cdot(-2) - 5)\cdot(-2) - 15]\cdot(-2) + 10\}\cdot(-2) + 8 = 0

2x^{4} - 5x^{3} - 15x^{2} + 10x + 8 = (x - 1)(x + 2)(ax^{2} + bx + c)

(x − 1)(x + 2) = x² + x − 2

(2x^{4} - 5x^{3} - 15x^{2} + 10x + 8) : (x^{2} + x - 2) = 2x^{2} - 7x - 4

$2x^{4}$	$+ 2x^{3}$	$- 4x^{2}$
0	$- 7x^{3}$	$- 11x^{2}$	$+ 10x$
	$- 7x^{3}$	$- 7x^{2}$	$+ 14x$
	0	$- 4x^{2}$	$- 4x$	$+ 8$
		$- 4x^{2}$	$- 4x$	$+ 8$
				$0$

(x^{2} + x - 2)(2x^{2} - 7x - 4) = 2x^{4} - 5x^{3} - 15x^{2} + 10x + 8

x = \frac{7 \pm \sqrt{(-7)^{2} - 4\cdot 2\cdot(-4)}}{2\cdot 2} = \frac{7 \pm \sqrt{49 + 32}}{4} = \frac{7 \pm 9}{4}

x_{1} = 4

x_{2} = -\tfrac{1}{2}

2x^{2} - 7x - 4 = 2(x - 4)\left(x + \tfrac{1}{2}\right) = (x - 4)(2x + 1)

2x^{4} - 5x^{3} - 15x^{2} + 10x + 8 = (x - 1)(x + 2)(x - 4)(2x + 1)

The same method, another polynomial

f(x) = 2x^{4} - 3x^{3} - 14x^{2} + 33x - 18 = \{[(2x - 3)x - 14]x + 33\}x - 18

x = 1 \;\Rightarrow\; \{[(2\cdot 1 - 3)\cdot 1 - 14]\cdot 1 + 33\}\cdot 1 - 18 = 0

x = 2 \;\Rightarrow\; \{[(2\cdot 2 - 3)\cdot 2 - 14]\cdot 2 + 33\}\cdot 2 - 18 = 0

2x^{4} - 3x^{3} - 14x^{2} + 33x - 18 = (x - 1)(x - 2)(ax^{2} + bx + c)

(x − 1)(x − 2) = x² − 3x + 2

(2x^{4} - 3x^{3} - 14x^{2} + 33x - 18) : (x^{2} - 3x + 2) = 2x^{2} + 3x - 9

$2x^{4}$	$- 6x^{3}$	$+ 4x^{2}$
0	$3x^{3}$	$- 18x^{2}$	$+ 33x$
	$3x^{3}$	$- 9x^{2}$	$+ 6x$
	0	$- 9x^{2}$	$+ 27x$	$- 18$
		$- 9x^{2}$	$+ 27x$	$- 18$
				$0$

(x^{2} - 3x + 2)(2x^{2} + 3x - 9) = 2x^{4} - 3x^{3} - 14x^{2} + 33x - 18

x = \frac{-3 \pm \sqrt{3^{2} - 4\cdot 2\cdot(-9)}}{2\cdot 2} = \frac{-3 \pm \sqrt{9 + 72}}{4} = \frac{-3 \pm 9}{4}

x_{1} = -3

x_{2} = \tfrac{3}{2}

2x^{2} + 3x - 9 = 2(x + 3)\left(x - \tfrac{3}{2}\right) = (x + 3)(2x - 3)

2x^{4} - 3x^{3} - 14x^{2} + 33x - 18 = (x - 1)(x - 2)(x + 3)(2x - 3)

Review exercises

Write f(x) = 6x³ − 5x² + 4x − 3 in nested form and calculate f(2).

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

f(2) = [(6\cdot 2 - 5)\cdot 2 + 4]\cdot 2 - 3 = 33

Without performing the entire division, determine the remainder.

(x^{4} - 2x^{3} + 3x^{2} - 4) : (x - 2)

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

f(2) = \{[(2 - 2)\cdot 2 + 3]\cdot 2 + 0\}\cdot 2 - 4 = 8

this is the remainder

Factorise the polynomial 6x⁴ + x³ − 25x² − 4x + 4.

6x^{4} + x^{3} - 25x^{2} - 4x + 4 = \{[(6x + 1)x - 25]x - 4\}x + 4

x = 2 \;\Rightarrow\; \{[(6\cdot 2 + 1)\cdot 2 - 25]\cdot 2 - 4\}\cdot 2 + 4 = 0

x = -2 \;\Rightarrow\; \{[(6\cdot(-2) + 1)\cdot(-2) - 25]\cdot(-2) - 4\}\cdot(-2) + 4 = 0

6x^{4} + x^{3} - 25x^{2} - 4x + 4 = (x - 2)(x + 2)(ax^{2} + bx + c)

(x − 2)(x + 2) = x² − 4

(6x^{4} + x^{3} - 25x^{2} - 4x + 4) : (x^{2} - 4) = 6x^{2} + x - 1

$6x^{4}$	$+ 0x^{3}$	$- 24x^{2}$
0	$x^{3}$	$- x^{2}$	$- 4x$
	$x^{3}$	$+ 0x^{2}$	$- 4x$
	0	$- x^{2}$	$+ 0x$	$+ 4$
		$- x^{2}$	$+ 0x$	$+ 4$
				$0$

(x^{2} - 4)(6x^{2} + x - 1) = 6x^{4} + x^{3} - 25x^{2} - 4x + 4

x = \frac{-1 \pm \sqrt{1^{2} - 4\cdot 6\cdot(-1)}}{2\cdot 6} = \frac{-1 \pm \sqrt{1 + 24}}{12} = \frac{-1 \pm 5}{12}

x_{1} = -\tfrac{1}{2}

x_{2} = \tfrac{1}{3}

6x^{2} + x - 1 = 6\left(x + \tfrac{1}{2}\right)\left(x - \tfrac{1}{3}\right) = 2\left(x + \tfrac{1}{2}\right)\cdot 3\left(x - \tfrac{1}{3}\right) = (2x + 1)(3x - 1)

6x^{4} + x^{3} - 25x^{2} - 4x + 4 = (x - 2)(x + 2)(2x + 1)(3x - 1)

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Frequently asked questions

How do you factor a fourth-degree polynomial?

Find a root by trial, divide by the corresponding binomial, and repeat on the cubic you get. A quadratic is left, and the quadratic formula factors it.

Why divide by the product of two binomials at once?

Because it is one division instead of two. If f(1) = 0 and f(−2) = 0, divide by (x − 1)(x + 2) = x² + x − 2 and a quadratic drops out immediately.

Which numbers are worth trying first?

The divisors of the constant term, so for 6x⁴ + x³ − 25x² − 4x + 4 those are ±1, ±2 and ±4. Only these can be whole-number roots.

How do you find the remainder without doing the division?

Evaluate the polynomial at a. The remainder on dividing x⁴ − 2x³ + 3x² − 4 by x − 2 is f(2) = 8, and no division is needed to get it.

A lesson from the PhiBoard knowledge base: phiboard.com/en/knowledge-base/from-zero-to-engineer/factoring-fourth-degree-polynomials. On the lesson page you will also find games and practice exercises - free and without an account.

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