How can you determine all of the zeroes (real and

2022-01-17 Answered
How can you determine all of the zeroes (real and imaginary) of the polynomial function:
P(x)=x4+2x39x210x24?
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Answers (2)

nick1337
Answered 2022-01-17 Author has 510 answers

Step 1
You can expect any factors to divide -24, the constant term over the lead coefficient. So check 1, 2, 3, 4, 6,12, 24 and their negatives. Turns out that 3 and -4 work, so you are down to a quadratic
P(x)=(x3)×(x+4)×(x2+x+2)
so the roots are
x={3, 4,17×i2, 1+7×i2}

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star233
Answered 2022-01-17 Author has 137 answers

Step 1
The basic trick is to try finding 3 numbers k,m,n which would enable us to write:
x4+2x39x210x24=(x2+x+k)2(mx+n)2
that is:
x4+2x3+(1+2km2)x2+2(kmn)x+(k2n2)=x4+2x39x210x24
The basic trick is to try finding 3 numbers k,m,n which would enable us to write:
1+2km2=92k2mn=10k2n2=24
Rearrange these equations and get
m2=2(k+5)mn=k+5n2=k2+24
From the 2 equations on the left we get
m2=2mn
and therefore one possible solution would be:
m=0k=5n2=49=72
Hence we get the following factorization of the given polynomial:
x4+2x39x210x24=(x2+x5)272==(x2+x+2)(x2+x12)
So, all the problem has now been reduced into finding the 4 roots of the 2 quadratic polynomials, which are
x1=12+72ix2=1272i
and
x3=12+72=3x4=1272=4

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