Sorry to update after almost a week, was really busy with other work. In this post, I shall talk about Polynomials. A polynomial is basically, a sequence of monomials, which is a term of product of powers of variables. Confused much ?
In simple terms, we just need to know about two polynomials, one we've already encountered, is a Quadratic, this is a polynomial of degree 2. The other we learn about in this post is a Cubic, which is a polynomial of degree 3.
We are expected to factorise cubics, aswell sketch them, and manipulate them algebraically, i.e. Factorise, Collect like terms and expand brackets. A cubic takes the form :
where a is not 0, and b,c,d are constants.
Cubics, can be factorised into 3 brackets, as Quadratics can be factorised using two. Where a cubic function crosses the x-axis is known as the roots. In terms of an equation, (the solutions), when the cubic is set equal to 0.
An sketch of some cubic function looks like :
Multiplying out Polynomials (Expanding Brackets)
If we have two polynomials (one a quadratic, the other linear) e.g.
(3x^2 + 2x + 8) (x + 5), if we had to expand them, its basically saying multiply each term in one bracket with the other term in the other bracket :
Multiply 3x^2 by the second bracket we get :
3x^3 + 15x^2
Multiply 2x by the second bracket we get :
2x^2 + 10x
Multiply 8 by the second bracket we get:
8x + 40
Collect like terms and simplify :
3x^3 + 15x^2 + 2x^2 + 10x + 8x + 40
= 3x^3 + 17x^2 + 18x + 40
Notice, its in the form of a cubic, where a = 3, b = 17, c = 18 and d = 40.
Though I don't think we need to be able to solve cubics, we can if the original quadratic we had can be factorised. If it can be factorised, we would have three roots, in this case the solutions we be :
x = -5, and x = one of the solutions of quadratic, x = other solution of quadratic