# Question ed07b

Oct 31, 2016

$f \left(x\right) = {x}^{3} - 4 {x}^{2} + 5 x - 2$

#### Explanation:

If the zero is $a$ and the multiplicity is $b$, then a factor of the polynomial will be ${\left(x - a\right)}^{b}$.

The factor for a zero of $2$ and multiplicity $1$ is ${\left(x - 2\right)}^{1}$.

The factor for a zero of $1$ and multiplicity $2$ is ${\left(x - 1\right)}^{2}$.

The polynomial is the product of the factors.

$f \left(x\right) = {\left(x - 2\right)}^{1} {\left(x - 1\right)}^{2}$.

Note: the sum of the exponents is $3$, which means the degree is $3$.

Rewriting:

$f \left(x\right) = \left(x - 2\right) \left(x - 1\right) \left(x - 1\right)$

Multiplying $\left(x - 1\right) \left(x - 1\right)$

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

Multiplying by $\left(x - 2\right)$

$f \left(x\right) = {x}^{3} - 2 {x}^{2} + x - 2 {x}^{2} + 4 x - 2$

Combining like terms

$f \left(x\right) = {x}^{3} - 4 {x}^{2} + 5 x - 2$