# 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 \(\normalsize{x}\) with arbitrary coefficients, such as \(\normalsize{p(x)=x^2-4x+3}\) or \(\normalsize{q(x)=x+2}.\) 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 \(\normalsize{1}\). There is an arithmetic with polynomials: they can be added, subtracted or multiplied just like numbers, so that for example \[\Large{(p+q)(x)=x^2-3x+5}\] \[\Large{(p-q)(x)=x^2-5x+1}\] \[\Large{(pq)(x)=x^3-2x^2-5x+6}.\]

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, Quadratic & Inverse Relations

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