One method if we know one factor involves juggling with coefficients, and is best demonstrated by example.
Consider the quartic
x^4 + 5x^3 + 7x^2 + 6 x + 8
It is easy to check that x = -2 gives a zero value for this function and hence (x + 2) is a factor.
The cubic factor can be found as follows.
x^4 + 5x^3 + 7x^2 + 6x + 8
= x^3(x + 2) + 3x^3 + 7x^2 + 6x + 8
= x^3(x + 2) + 3x^2 (x + 2) + x^2 + 6x + 8
= (x^3 + 3x^2) (x + 2) + x(x + 2) + 4x + 8
= (x^3 + 3x^2 + x) (x + 2) + 4(x + 2)
= (x^3 + 3x^2 + x + 4) (x + 2)
Hence the cubic factor is (x^3 + 3x^2 + x + 4)
Worked Examples
Find the missing cubic expression here:
x^4 - 3x^3 + 5x^2 + x - 4 = (x - 1) ( \text{cubic} )
x^4 - 3x^3 + 5x^2 + x - 4
= x^3 (x - 1) - 2x^3 + 5x^2 + x - 4
= x^3 (x - 1) - 2x^2 (x - 1) + 3x^2 + x - 4
= (x^3 - 2x^2) (x - 1) + 3x (x - 1) + 4x - 4
= (x^3 - 2x^2 + 3x) (x - 1) + 4(x - 1)
= (x^3 - 2x^2 + 3x + 4) (x - 1)
So the missing expression is x^3 - 2x^2 + 3x + 4.
The second method is sometimes called 'long division of polynomials'.
The statement
x^4 + 5x^3 + 7x^2 + 6x + 8 = (x + 2) ( \text{cubic} )
can be written instead in this form
\frac{x^4 + 5x^3 + 6x + 8}{x + 2} = \text{cubic polynomial}
The cubic polynomial is thus the result of dividing the quartic by (x + 2).
This division can be accomplished in a manner very similar to long division of numbers as we see below.
| x^3 | + | 3x^2 | + | x | + | 4 | |||
| x+2 | x^4 | + | 5x^3 | + | 7x^2 | + | 6x | + | 8 |
| x^4 | + | 2x^3 | |||||||
| 3x^3 | + | 7x^2 | |||||||
| 3x^3 | + | 6x^2 | |||||||
| x^2 | + | 6x | |||||||
| x^2 | + | 2x | |||||||
| 4x | + | 8 | |||||||
| 4x | + | 8 | |||||||
| 0 |
So
x^4 + 5x^5 + 7x^2 + 6x + 8 = (x + 2) (x^3 + 3x^2 + x + 4)
This is same answer is obtained as on the previous page by the 'juggling' method.
Evaluate the quotient \frac{x^4 - 5x^3 - 2x^2 + 25x - 3}{x - 3}
| x^3 | - | 2x^2 | - | 8x | + | 1 | |||
| x-3 | x^4 | - | 5x^3 | - | 2x^2 | + | 25x | - | 3 |
| x^4 | - | 3x^3 | |||||||
| - | 2x^3 | - | 2x^2 | ||||||
| - | 2x^3 | + | 6x^2 | ||||||
| - | 8x^2 | + | 25x | ||||||
| - | 8x^2 | + | 24x | ||||||
| x | - | 3 | |||||||
| x | - | 3 | |||||||
| 0 | |||||||||
We can use these methods to help solve quartics or even higher order polynomials but in reality for practical work, numerical and graphical approaches are both easier and more appropriate to the problem. Here though is one algebraic approach to solving a quartic equation.
Factorise the quartic x^4 - 4x^3 - 7x^2 + 34x - 24 as fully as possible and hence solve the equation
x^4 - 4x^3 - 7x^2 + 34x - 24 = 0
x = 1 gives zero value for the quartic, so (x – 1) is a factor.
Long division gives
x^4 - 4x^3 - 7x^2 + 34x - 24 = (x - 1) (x^3 - 3x^2 - 10x + 24)
Now factorise the cubic.
Since x = 2 is a zero of the cubic, (x - 2) is a factor.
Long division now gives
x^3 - 3x^2 - 10x + 24 = (x - 2) (x^2 - x - 12)
The quadratic x^2 - x - 12 factorises easily to give (x - 4) (x + 3).
The full factorisation of the original quartic is therefore
(x - 1) (x - 2) (x - 4) (x + 3)
and the solutions to the equation are thus
x = −3, 1, 2 and 4
Solve the equation x^4 + 3x^3 - 11x^2 - 19x - 6 = 0.
First, factorise the quartic as far as possible.
x = -1 gives a zero value for the quartic so (x + 1) is a factor.
By long division,
x^4 + 3x^3 - 11x^2 - 19x – 6 = (x + 1)(x^3 + 2x^2 - 13x - 6)
Also x = 3 gives a zero value for the cubic so (x - 3) is a factor and
x^3 + 2x^2 - 13x - 6 = (x - 3)(x^2 + 5x + 2)
The quadratic x^2 + 5x + 2 has no easy linear factors, but the equation x^2 + 5x + 2 = 0 does have solutions, namely -4.56 and -0.438.
So the quartic thus factorises to (x + 1)(x - 3)(x^2 + 5x + 2) giving solutions
x = -4.56, -1, -0.438 and 3
Exercises
