Find All Complex Number Solutions 30x^4-97x^3+50x^2-4x-1=0

Problem

30*x4−97*x3+50*x2−4*x−1=0

Solution

1. Test for rational roots using the Rational Root Theorem, checking factors of the constant term −1 over factors of the leading coefficient 30

2. Substitute x=3 into the polynomial: 30(81) - 97(27) + 50(9) - 4(3) - 1 = 2430 - 2619 + 450 - 12 - 1 = 248$. Since this is not zero, try a different value.

3. Substitute x=1/2 into the polynomial: 30*(1/16)−97*(1/8)+50*(1/4)−4*(1/2)−1=15/8−97/8+100/8−16/8−8/8=(−6)/8≠0

4. Substitute x=1/3 into the polynomial: 30*(1/81)−97*(1/27)+50*(1/9)−4*(1/3)−1=10/27−97/27+150/27−36/27−27/27=0 Thus, (3*x−1) is a factor.

5. Divide the polynomial by (3*x−1) using long division or synthetic division to obtain the quotient 10*x3−29*x2+7*x+1

6. Test for rational roots in the new cubic polynomial 10*x3−29*x2+7*x+1 Substituting x=1/5 gives 10*(1/125)−29*(1/25)+7*(1/5)+1=2/25−29/25+35/25+25/25≠0

7. Substitute x=(−1)/10 into the cubic: 10*((−1)/1000)−29*(1/100)+7*((−1)/10)+1=(−1)/100−29/100−70/100+100/100=0 Thus, (10*x+1) is a factor.

8. Divide the cubic 10*x3−29*x2+7*x+1 by (10*x+1) to obtain the quadratic x2−3*x+1

9. Apply the quadratic formula to x2−3*x+1=0 where a=1,b=−3,c=1

10. Calculate the roots: x=(−(−3)±√(,(−3)2−4*(1)*(1)))/(2*(1))=(3±√(,5))/2

Final Answer

x={1/3,−1/10,(3+√(,5))/2,(3−√(,5))/2}

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