# How do you find the polynomial function with roots 1, 7, and -3 of multiplicity 2?

Aug 10, 2015

$f \left(x\right) = 2 \left(x - 1\right) \left(x - 7\right) \left(x + 3\right) = 2 {x}^{3} - 5 {x}^{2} - 17 x + 21$

#### Explanation:

If the roots are 1,7,-3 then in factored form the polynomial function will be:

$f \left(x\right) = A \left(x - 1\right) \left(x - 7\right) \left(x + 3\right)$

Repeat the roots to get the required multiplicity:

$f \left(x\right) = \left(x - 1\right) \left(x - 7\right) \left(x + 3\right) \left(x - 1\right) \left(x - 7\right) \left(x + 3\right)$

Aug 11, 2015

The simplest polynomial with roots $1$, $7$ and $- 3$, each with multiplicity $2$ is:

$f \left(x\right) = {\left(x - 1\right)}^{2} {\left(x - 7\right)}^{2} {\left(x + 3\right)}^{2}$

$= {x}^{6} - 10 {x}^{5} - 9 {x}^{4} + 212 {x}^{3} + 79 {x}^{2} - 714 x + 441$

#### Explanation:

Any polynomial with these roots with at least these multiplicities will be a multiple of $f \left(x\right)$, where...

$f \left(x\right) = {\left(x - 1\right)}^{2} {\left(x - 7\right)}^{2} {\left(x + 3\right)}^{2}$

$= {\left({x}^{3} - 5 {x}^{2} - 17 x + 21\right)}^{2}$

$= {x}^{6} - 10 {x}^{5} - 9 {x}^{4} + 212 {x}^{3} + 79 {x}^{2} - 714 x + 441$

...at least I think I've multiplied this correctly.

Let's check $f \left(2\right)$ :

${2}^{6} - 10 \cdot {2}^{5} - 9 \cdot {2}^{4} + 212 \cdot {2}^{3} + 79 \cdot {2}^{2} - 714 \cdot 2 + 441$

$= 64 - 320 - 144 + 1696 + 316 - 1428 + 441 = 625$

${\left(\left(2 - 1\right) \left(2 - 7\right) \left(2 + 3\right)\right)}^{2} = {\left(1 \cdot - 5 \cdot 5\right)}^{2} = {\left(- 25\right)}^{2} = 625$