SOLUTION: 1.) What is a polynomial function in standard form of degree 4 with zeros: -1, 2, 5i? 2.) Given f(x)=5x+1 Find f^-1(x) & verify the answer is correct by showing that (fo

Algebra ->  Complex Numbers Imaginary Numbers Solvers and Lesson -> SOLUTION: 1.) What is a polynomial function in standard form of degree 4 with zeros: -1, 2, 5i? 2.) Given f(x)=5x+1 Find f^-1(x) & verify the answer is correct by showing that (fo      Log On


   

Question 117426: 1.) What is a polynomial function in standard form of degree 4 with zeros: -1, 2, 5i?
2.) Given f(x)=5x+1
Find f^-1(x) & verify the answer is correct by showing that (fof^-1)(x)=x
3.) solve: a) log (x+4) - log 2x=0
b) 2ln x + 5 = 7
4.) Find all the zeros of the polynomial function: P(x)=2x^4+3x^3+16x^2+27x-18 then write the P(x) as a product of its linear factors.

Answer by psbhowmick(878) About Me  (Show Source):
You can put this solution on YOUR website!
1.) For any polynomial equation with real coefficients, complex roots occur in conjugate pairs.
So, if 5i i.e. (0 + 5i) is a root then (0 - 5i) is the other root.
Hence, the 4 zeros are x = -1, 2, 5i, - 5i
Hence the polynomial is
%28x+-+%28-1%29%29%28x+-+2%29%28x+-+5i%29%28x+-+%28-5i%29%29
= %28x+%2B+1%29%28x+-+2%29%28x+-+5i%29%28x+%2B5i%29
= %28x+%2B+1%29%28x+-+2%29%28x%5E2+-+25i%5E2%29
= %28x+%2B+1%29%28x+-+2%29%28x%5E2+%2B+25%29
= %28x%5E2+-+x+-+2%29%28x%5E2+%2B+25%29
= x%5E4+-+x%5E3+%2B+23x%5E2+-+25x+-+50
2.)
f(x) = 5x + 1
5x = f(x) - 1
x+=+%28f%28x%29-1%29%2F5
Hence, f%5E%28-1%29%28x%29+=+%28x+-+1%29%2F5

Verification:
f%28f%5E%28-1%29%28x%29%29+=+5f%5E%28-1%29%28x%29+%2B+1+=+5%28x-1%29%2F5+%2B+1+=+x
3.a) log%2810%2Cx%2B4%29+-+log+%2810%2C2x%29+=+0
log%2810%2C%28x%2B4%29%2F2x%29+=+0
%28x%2B4%29%2F2x+=+1
%28x%2B4%29+=+2x
x+=+4
3.b) 2log%28e%2Cx%29+%2B+5+=+7
2log%28e%2Cx%29+=+7+-+5+=+2
log%28e%2Cx%29+=+1
x+=+e
4.) Try yourself!