You use the rational root theorem and the quadratic formula to find the roots.
f(x)=2x4−5x3+3x2+4x−6
According to the rational root theorem, the rational roots of f(x)=0 must all be of the form pq, with p a divisor of the constant term −6 and q a divisor of the coefficient 2 of x4.
So the only possible rational roots are:
±1,±2,±3,±6,±12,±32
By trial and error, we find:
f(−1)=2+5+3−4−6=0
f(32)=2(32)4−5(32)3+3(33)2+4(32)−6
=2(8116)−5(278)+3(94)+4(32)−6=818−1358+274+122−6=81−135+54+48−488=0
So x=−1 and x=32 are roots of f(x)=0, and (x+1) and (x−32) are factors of f(x).
Divide f(x) by (x+1)(x−32)=x2+52x+32 to find:
2x4−5x3+3x2+4x−6x2+52x+32=2x2−4x+4=2(x2−2x+2)
f(x)=2(x+1)(x−32)(x2−2x+2)=(x+1)(2x−3)(x2−2x+2)
Using the quadratic formula, we find that the roots of x2−2x+2=0 are:
x=−(−2)±√(−2)2−4×1×22×1=2±√4−82=2±√−42=2±2i2=1±i
The zeroes of f(x)=2x4−5x3+3x2+4x−6 are x=−1,x=32,x=1−i, and x=1+i.
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