Asked by Tan
Solve the equation 4x^3 + 32x^2 + 79x + 60 =0 given that the root is equal to the sum of the other two roots.
(Hint: set roots as alpha, beta, gamma)
(Hint: set roots as alpha, beta, gamma)
Answers
Answered by
Reiny
Using the factor theorem , I tried
x = ±1, ±2, ±4, and it worked if x = -4
so by synthetic division:
4x^3 + 32x^2 + 79x + 60 = (x-4)(4x^2 + 16x + 15)
= (x-4)(2x+5)(2x+3)
(x-4)(2x+5)(2x+3)=0
x = -4, or x = -5/2, or x = -3/2
and sure enough -5/2 + (-3/2)
= -8/2
= -4
x = ±1, ±2, ±4, and it worked if x = -4
so by synthetic division:
4x^3 + 32x^2 + 79x + 60 = (x-4)(4x^2 + 16x + 15)
= (x-4)(2x+5)(2x+3)
(x-4)(2x+5)(2x+3)=0
x = -4, or x = -5/2, or x = -3/2
and sure enough -5/2 + (-3/2)
= -8/2
= -4
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