# Factors of Quadratic Polynomials and Zeroes

- review basics of polynomial arithmetic
- relate zeroes and factors of quadratics via Descartes theorem
- give a proof of Descartes theorem.

## Polynomials and their arithmetic

A**polynomial**is a general algebraic expression made from powers of \(\normalsize{x}\) mixed with arbitrary

**coefficients**, such as\[\normalsize{p(x)=x^2-4x+3} \quad\text{ or } \quad \normalsize{q(x)=x+2}.\]The

**degree**of a polynomial is the highest power of \(\normalsize{x}\) that appears, so that \(\normalsize{p}\) has degree \(\normalsize{2}\) and \(\normalsize{q}\) has degree \(\normalsize{1}\). The coefficients of \(\normalsize{p}\) are \(\normalsize{1,-4}\) and \(\normalsize{3}\) corresponding to \(\normalsize{x^2,x}\) and \(\normalsize{1}\) respectively.There is an arithmetic with polynomials: addition or subtraction is just addition or subtraction of the corresponding coefficients. For example\[\Large{(p+q)(x)=x^2-3x+5}\] \[\Large{(p-q)(x)=x^2-5x+1}.\]Multiplication by a number is also performed on all of the coefficients independently\[\Large{(2p)(x) = 2(x^2 – 4x + 3) = 2x^2-8x+6}.\]Multiplication by \(\normalsize{x}\) increases the exponent of all the \(\normalsize{x}\)’s by one.\[\Large{(xp)(x) = x(x^2 – 4x + 3) = x^3-4x^2+3x}.\]Finally, multiplication by a polynomial combines these operations.\[\Large{\begin{align*} (qp)(x) & =(x+2)p(x) \\ & = (xp)(x) + (2p)(x) \\ & = (x^3-4x^2+3x) + (2x^2-8x+6) \\ & = x^3-2x^2-5x+6.\end{align*}}\]

Q1(E): If \(\normalsize{r(x)=x+3}\) and \(\normalsize{s(x)=2x-1}\) then determine the polynomials i) \(\normalsize{r+s}\) ii) \(\normalsize{r-s}\) and iii) \(\normalsize{rs}\).Q2(E): If \(\normalsize{u(x)=x^2+x+1}\) and \(\normalsize{v(x)=x^3-1}\) then determine the polynomials i) \(\normalsize{u+v}\) ii) \(\normalsize{u-v}\) and iii) \(\normalsize{uv}\).

## Factoring polynomials

**multiple**of \(\normalsize{x+1}\), and that equivalently \(\normalsize{x+1}\) is a

**factor**of \(\normalsize{x^2-1}\).

\[\Large{30x^2-28x-240}.\]Q3(M): I have multiplied two linear factors together to get the quadratic polynomialWhat were my two linear polynomials?

## Factoring challenge

## Zeroes of a quadratic polynomial

**zero**of \(\normalsize{p(x)}\) is a number \(\normalsize{r}\) with the property that \(\normalsize{p(r)=0}\). In other words, we are solving the equation \(\normalsize{r^2-12r+35=0}\). The basic technique for solving this kind of quadratic equation goes back to the ancient Hindus.

\(\normalsize{x}\) | \(\normalsize{p(x)}\) |
---|---|

-3 | 80 |

-2 | 63 |

-1 | 48 |

0 | 35 |

1 | 24 |

2 | 15 |

3 | 8 |

4 | 3 |

5 | 0 |

6 | -1 |

7 | 0 |

8 | 3 |

9 | 8 |

10 | 15 |

## The connection with factors

**linear factor**\(\normalsize{(x-r)}\) when \(\normalsize{p(x)=(x-r)q(x)}\) for some other polynomial \(\normalsize{q(x)}\). Descartes realized that the zeroes of \(\normalsize{p(x)}\) were intimately connected with its linear factors. Note that we can write

Q4(E): What are the linear factors of \(\normalsize{x^2+8x+15}\)?

**Theorem (Descartes’ Factor Theorem)**

*If \(\normalsize{p(x)}\) is a polynomial, then \(\normalsize{p(x)}\) has a zero \(\normalsize{r}\) precisely when \(\normalsize{(x-r)}\) is a factor of \(\normalsize{p(x)}\)*.

- if \(\normalsize{p(x)}\) has a zero \(\normalsize{r}\) then \(\normalsize{(x-r)}\) is a factor, and
- if \(\normalsize{(x-r)}\) is a factor then \(\normalsize{p(x)}\) has a zero \(\normalsize{r}\).

## Examples with Descartes’ Factor Theorem

Q5(M): What are the linear factors of \(\normalsize{x^2+\frac{7}{2}x+3}\)?

Q6(C): How many linear factors does the polynomial have whose graph is shown in the image to this step?

## A proof of Descartes theorem (advanced)

**Proof**: If \(\normalsize{(x-r)}\) is a factor of \(\normalsize{p(x)=ax^2+bx+c}\), it means that \(\normalsize{p(x)=(x-r)q(x)}\) for some other polynomial \(\normalsize{q(x)}\). In that case if we substitute \(\normalsize{r}\) we have \(\normalsize{p(r)=(r-r)q(r)}\) which means that \(\normalsize{p(r)=0}\) so that yes, \(\normalsize{r}\) is a zero of \(\normalsize{p(x)}\).

## Answers

A1.i) The sum is \(\normalsize{(r+s)(x)=x+3+2x-1=3x+2}\).ii) The difference is \(\normalsize{(r-s)(x)=x+3-(2x-1)=-x+4}\).iii) The product is \(\normalsize{(rs)(x)=(x+3)\times(2x-1)=2x^2+5x-3}\).A2.i) The sum is \(\normalsize{(u+v)(x)=x^2+x+1+x^3-1=x^3+x^2+x}\).ii) The difference is \(\normalsize{(u-v)(x)=x^2+x+1-(x^3-1)=-x^3+x^2+x+2}\).iii) The product is \(\normalsize{\begin{align}(uv)(x)&=(x^2+x+1)\times(x^3-1)\\&=x^5-x^2+x^4-x+x^3-1 \\&=x^5+x^4+x^3-x^2-x-1. \end{align}}\)\[\Large{30x^2-28x-240=(5x+12)(6x-20)}.\]A3.The factorization is:How could we get that? We need to find two numbers that multiply to \(\normalsize{30}\), and two other numbers that multiply to \(\normalsize{-240}\) such that the sum of products of two of them with the other two is \(\normalsize{-28}\). It certainly appears that trial and error are needed!A4.The linear factors of \(\normalsize{x^2+8x+15}\) are \((x+3)\) and \((x+5)\).A5.The linear factors of \(\normalsize{x^2+\frac{7}{2} x+3}\) are \(\normalsize{(x+\frac{3}{2})}\) and \(\normalsize{(x+2)}\).\[\Large{p(x)= (x+3)(x+2)(x+1)x(x-1)(x-2)(x-3)}.\]A6.The polynomial visually has zeros at \(\normalsize{x=-3,-2,-1,0,1,2}\) and \(\normalsize{3}\). So according to Descartes’ theorem, it will have factors \(\normalsize{(x+3),(x+2),(x+1),x,(x-1),(x-2)}\) and \(\normalsize{(x-3)}\). So a good guess for the polynomial might be:

#### Maths for Humans: Linear and Quadratic Relations

## Our purpose is to transform access to education.

We offer a diverse selection of courses from leading universities and cultural institutions from around the world. These are delivered one step at a time, and are accessible on mobile, tablet and desktop, so you can fit learning around your life.

We believe learning should be an enjoyable, social experience, so our courses offer the opportunity to discuss what you’re learning with others as you go, helping you make fresh discoveries and form new ideas.

You can unlock new opportunities with unlimited access to hundreds of online short courses for a year by subscribing to our Unlimited package. Build your knowledge with top universities and organisations.

Learn more about how FutureLearn is transforming access to education