Arthur Pratt

2021-12-15

Find a degree 3 polynomial with the specified number of zeros that also satisfies the given condition.

−5, 2, 1; f(3) = 64

f(x) =

veiga34

Beginner2021-12-16Added 32 answers

Step 1

Here, we apply the linear factors to the provided zeroes

see below the calculation

Step 2

Given zeros are −5,2,1

then the polynomial's linear dividing factor 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 $f\left(x\right)=\frac{16}{3}(x+5)(x-2)(x-1)$

$=\frac{16}{3}({x}^{3}+2{x}^{2}-13x+10)$

Pansdorfp6

Beginner2021-12-17Added 27 answers

Given zeroes -5,2,1

Equation of polynomial

f(x)=k(x-(-5))(x-2)(x-1)

=k(x+5)(x-2)(x-1)

$=k({x}^{2}-2x+5x-10)(x-1)$

$=k({x}^{2}+3x-10)(x-1)$

$=k({x}^{3}-{x}^{2}+3{x}^{2}-3x-10x+10)$

$=k({x}^{3}+2{x}^{2}-13x+10)$

$\therefore f\left(x\right)=k({x}^{3}+2{x}^{2}-13x+10)$

Given f(3)=64

$\therefore k({3}^{3}+2x{3}^{2}-13x3+10)=64$

$\Rightarrow k(27+18-39+10)=64$

$\Rightarrow 16k=64$

$\Rightarrow k=\frac{64}{16}-4$

$\therefore f\left(x\right)=4({x}^{3}+2{x}^{2}-13x+10)$

Equation of polynomial

f(x)=k(x-(-5))(x-2)(x-1)

=k(x+5)(x-2)(x-1)

Given f(3)=64

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