 segnverd3a

2022-01-29

How do you write a polynomial function in standard form with the zeroes x=3, -2, 1? Jude Carpenter

Expert

f(x)=(x-3)(x+2)(x-1)
$=\left({x}^{2}-x-6\right)\left(x-1\right)$
$=\left({x}^{2}-x-6\right)x-\left({x}^{2}-x-6\right)$
$={x}^{3}-2{x}^{2}-5x+6$ sjkuzy5

Expert

First, we should establish what it means to be a zero. If the function is "zero" at those values, that means that y= 0 at those specific values of x.
Think about what a factored function looks like. It usually is something like
f(x) = (x+2)(x-3) or something like that.
The zeros for the previous function are where (x - 2) = 0 or where (x + 3) = 0. Now we use this general idea with your given zeros.
f(x)=(x+2)(x-3)(x-1)
To make this function in standard form, we need to multiply it all out. I prefer to work with the two left parts first. FOIL them out to get:
$f\left(x\right)=\left({x}^{2}-3x-2x-6\right)\left(x-1\right)$
Now multiply those together to get
$f\left(x\right)={x}^{3}-{x}^{2}-{x}^{2}-x+6x+6$
Combining like terms again and we find the polynomial in standard form:
$f\left(x\right)={x}^{3}-2{x}^{2}-5x+6$

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