ASKED THESE QUESTIONS TWICE. NOBODY HAS BEEN ABLE TO SOLVE ON CHEGG. ANSWER FORM

Question

ASKED THESE QUESTIONS TWICE. NOBODY HAS BEEN ABLE TO SOLVE ON CHEGG. ANSWER FORMAT BELOW.

Degree Zeros Solution Point 4 -1, 2, V2 f(1) = 18 Need Help? LLRead Subomit Answer Save Progres Paice Another Version Practice Another Version 2. -18.33 points LarPCalc10 2.5.050 Find the polynomial function f with real coefficients that has the given deg Degree Zeros Solution Point -2, 1-y2 -1) =-36 Need Help? Read It 3.0 -8.33 points LarPCalc10 2.5 055 Use the given zero to find all the zeros of the function. (Enter your answers zero in your answer.) Functionn Zero Need Help? Read It

Explanation / Answer

multiple questions posted.please post each question seperately

1)

given -1,2,2 i are zeros. since all the quotients are real -2 i is also a zero

factors of polynomial are (x+1),(x-2),(x-2 i),(x+2 i)

polynomial is of the form f(x)=k (x+1)(x-2)(x-2 i)(x+2 i)

f(x)=k (x2-x-2)(x2+2 )

given f(1)=18

=>k (12-1-2)(12+2 )=18

=>-6k =18

=>k=-3

=>f(x)=-3(x2-x-2)(x2+2 )

=>f(x)=-3x4+3x3+6x+12

so the polynomial is f(x)=-3x4+3x3+6x+12

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2)

given -2,1-2 i are zeros. since all the quotients are real 1+2 i is also a zero

factors of polynomial are (x+2),(x-1+2 i),(x-1-2 i)

polynomial is of the form f(x)=k  (x+2)(x-1+2 i)(x-1-2 i)

f(x)=k  (x+2)((x-1)2+2)

f(-1)=-36

=>k  (-1+2)((-1-1)2+2)=-36

=>k(1)(4+2)=-36

=>k=-6

=> f(x)=-6 (x+2)((x-1)2+2)

=> f(x)=-6(x+2)(x2-2x+3)

=> f(x)=-6(x3-2x2+3x+2x2-4x+6)

=> f(x)=-6x3+6x-36

so the polynomial is  f(x)=-6x3+6x-36

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