SOLUTION: hello. please help me with these.. ( ' means exponet # after) Solve each equation. 1. 2x'3 – 5x'2 – 3x + 2 = 2 2. x'3 + 64 = 0 3. x4 – 8x'2 – 9 = 0

Algebra ->  Rational-functions -> SOLUTION: hello. please help me with these.. ( ' means exponet # after) Solve each equation. 1. 2x'3 – 5x'2 – 3x + 2 = 2 2. x'3 + 64 = 0 3. x4 – 8x'2 – 9 = 0       Log On


   

Question 174369: hello. please help me with these.. ( ' means exponet # after)

Solve each equation.
1. 2x'3 – 5x'2 – 3x + 2 = 2

2. x'3 + 64 = 0

3. x4 – 8x'2 – 9 = 0


Answer by Earlsdon(6294) About Me  (Show Source):
You can put this solution on YOUR website!
Solve:
1) 2x%5E3-5x%5E2-3x%2B2+=+2 Subtract 2 from both sides.
2x%5E3-5x%5E2-3x+=+0 Factor an x from the left side.
x%282x%5E2-5x-3%29+=+0 Factor the the parentheses.
x%282x%2B1%29%28x-3%29+=+0 Apply the zero product rule.
x+=+0 or 2x%2B1+=+0 or x-3+=+0, so...
highlight%28x+=+0%29
highlight%28x+=+-1%2F2%29
highlight%28x+=+3%29
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2) x%5E3%2B64+=+0 This is a sum of cubes.
%28x%29%5E3%2B%284%29%5E3+=+0 This can be solved thus:A%5E3%2BB%5E3+=+%28A%2BB%29%28A%5E2-AB%2BB%5E2%29
%28x%2B4%29%28x%5E2-4x%2B16%29+=+0 Applying the zero product rule:
x%2B4+=+0 or x%5E2-4x%2B16+=+0
highlight%28x+=+-4%29 or
x+=+%28-%28-4%29%2B-sqrt%28%284%29%5E2-4%281%29%2816%29%29%29%2F2%281%29
x+=+%284%2B-sqrt%28-48%29%29%2F2
x+=+%284%2B-sqrt%2816%2A%28-3%29%29%29%2F2
x+=+2%2B2sqrt%28-3%29 or x+=+2-2sqrt%28-3%29 and these can be expressed in terms of i, as:
highlight%28x+=+2%2B2sqrt%283%29i%29 or highlight%28x+=+2-2sqrt%283%29i%29
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3) x%5E4-8x%5E2-9+=+0 Rewrite in terms of x%5E2
%28x%5E2%29%5E2-8%28x%5E2%29-9+=+0 Factor this.
%28x%5E2%2B1%29%28x%5E2-9%29+=+0 Apply the zero product rule:
x%5E2%2B1+=+0 or x%5E2-9+=+0, so...
x%5E2+=+-1 Take the square root of both sides.
x+=+sqrt%28-1%29 or x+=+-sqrt%28-1%29 or, in terms of i,
highlight%28x+=+i%29 or highlight%28x+=+-i%29
x%5E2-9+=+0
x%5E2+=+9 Take the square root of both sides.
highlight%28x+=+3%29 or highlight%28x+=+-3%29