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

Arthur Pratt

Arthur Pratt

Answered question

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) =

Answer & Explanation

veiga34

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=163 
then required 3 degree polynomial is f(x)=163(x+5)(x2)(x1) 
=163(x3+2x213x+10)

Pansdorfp6

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(x22x+5x10)(x1)
=k(x2+3x10)(x1)
=k(x3x2+3x23x10x+10)
=k(x3+2x213x+10)
f(x)=k(x3+2x213x+10)
Given f(3)=64
k(33+2x3213x3+10)=64
k(27+1839+10)=64
16k=64
k=64164
f(x)=4(x3+2x213x+10)

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