How can you determine all of the zeroes (real and imaginary) of the polynomial function: P(x)=x4+2x3−9x2−10x−24?

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How can you determine all of the zeroes (real and imaginary) of the polynomial function:  P(x)=x4+2x3−9x2−10x−24?

nick1337 · Accepted Answer

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)=(x−3)×(x+4)×(x2+x+2) so the roots are x={3, −4,−1−7×i2, −1+7×i2}

star233 · Accepted Answer

Step 1 The basic trick is to try finding 3 numbers k,m,n which would enable us to write: x4+2x3−9x2−10x−24=(x2+x+k)2−(mx+n)2 that is: x4+2x3+(1+2k−m2)x2+2(k−mn)x+(k2−n2)=x4+2x3−9x2−10x−24 The basic trick is to try finding 3 numbers k,m,n which would enable us to write: 1+2k−m2=−92k−2mn=−10k2−n2=−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+2x3−9x2−10x−24=(x2+x−5)2−72==(x2+x+2)(x2+x−12) So, all the problem has now been reduced into finding the 4 roots of the 2 quadratic polynomials, which are x1=−12+72ix2=−12−72i and x3=−12+72=3x4=−12−72=−4

How can you determine all of the zeroes (real and

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