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Answer 1 · 2016-10-31T00:40:59

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

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

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

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

The polynomial is the product of the factors.

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

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

Rewriting:

$f (x) = (x - 2) (x - 1) (x - 1)$ $f(x)=(x-2)(x-1)(x-1)$

Multiplying $(x - 1) (x - 1)$ $(x-1)(x-1)$

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

Multiplying by $(x - 2)$ $(x-2)$

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

Combining like terms#

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