Which two techniques does factorising a cubic combine?

The factor theorem , to find one linear factor, and polynomial division , to get the rest. Neither is enough alone: the theorem gives a factor but not the quotient, and division needs a divisor before it can start.

Can the same method be used on a polynomial of degree higher than three?

Yes, because each linear factor found reduces the degree by one. A quartic simply needs the find-a-factor-then-divide cycle carried out twice before a quadratic is left.

International A Level (IAL)MathsEdexcelPure 2FlashcardsAlgebra & FunctionsPolynomials

Polynomials (Edexcel International A Level (IAL) Maths: Pure 2): Flashcards

Exam code: YMA01

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Which two techniques does factorising a cubic combine?

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Which two techniques does factorising a cubic combine?
The factor theorem, to find one linear factor, and polynomial division, to get the rest.
Neither is enough alone: the theorem gives a factor but not the quotient, and division needs a divisor before it can start.
What is the goal when fully factorising a polynomial?
To write it as a product of linear factors, taken as far as it will go.
For example $x^{3} + 4 x^{2} - 11 x - 30 = (x + 2) (x - 3) (x + 5)$ .
What is the first move in factorising a cubic $f (x)$ ?
Find a value $p$ for which $f (p) = 0$ .
Until you have one factor there is nothing to divide by, so this has to come first.
True or False?
Every cubic can be written as a product of three linear factors.
False.
If the quadratic left after dividing does not factorise, the answer stops at one linear factor times a quadratic.
For example $2 x^{3} + 3 x^{2} + 10 x - 6 = (2 x - 1) (x^{2} + 2 x + 6)$ .
Factorising $x^{3} + 6 x^{2} - 9 x - 14$ : since $f (- 1) = 0$ , dividing by $(x + 1)$ leaves $x^{2} + 5 x - 14$ , which factorises as:
$(x +______) (x -______)$
$(x + 7) (x - 2)$
So the full factorisation is $x^{3} + 6 x^{2} - 9 x - 14 = (x + 1) (x + 7) (x - 2)$ .
Can the same method be used on a polynomial of degree higher than three?
Yes, because each linear factor found reduces the degree by one.
A quartic simply needs the find-a-factor-then-divide cycle carried out twice before a quadratic is left.

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