# Find a polynomial f(x) of degree 3 that has the

Find a polynomial f(x) of degree 3 that has the indicated zeros and satisfies the given condition.
−5, 2, 1; f(3) = 64
f(x) =
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veiga34
Step 1
here we use the linear factors for the zeroes given
see below the calculation
Step 2
Given zeros are −5,2,1
then the linear factor that divide the polynomial f(x) are (x+5),
(x-2),(x-1)
then f(x)=m(x+5)(x-2)(x-1) but to satisfy the another conditions that f(3)=64
we see that f(3)=m(3+5)(3-2)(3-1)=12m
then 12m=64 we get $m=\frac{16}{3}$
then required 3 degree polynomial is $f\left(x\right)=\frac{16}{3}\left(x+5\right)\left(x-2\right)\left(x-1\right)$
$=\frac{16}{3}\left({x}^{3}+2{x}^{2}-13x+10\right)$
###### Not exactly what you’re looking for?
Pansdorfp6
Given zeroes -5,2,1
Equation of polynomial
f(x)=k(x-(-5))(x-2)(x-1)
=k(x+5)(x-2)(x-1)
$=k\left({x}^{2}-2x+5x-10\right)\left(x-1\right)$
$=k\left({x}^{2}+3x-10\right)\left(x-1\right)$
$=k\left({x}^{3}-{x}^{2}+3{x}^{2}-3x-10x+10\right)$
$=k\left({x}^{3}+2{x}^{2}-13x+10\right)$
$\therefore f\left(x\right)=k\left({x}^{3}+2{x}^{2}-13x+10\right)$
Given f(3)=64
$\therefore k\left({3}^{3}+2x{3}^{2}-13x3+10\right)=64$
$⇒k\left(27+18-39+10\right)=64$
$⇒16k=64$
$⇒k=\frac{64}{16}-4$
$\therefore f\left(x\right)=4\left({x}^{3}+2{x}^{2}-13x+10\right)$