SOLUTION: Factor completely using integer coefficients: e) x^4+8x^3-2x^2-16x f) 2x^6-20x^4-16x^3+160x j) x^4-x^3-x+1

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: Factor completely using integer coefficients: e) x^4+8x^3-2x^2-16x f) 2x^6-20x^4-16x^3+160x j) x^4-x^3-x+1      Log On


   



Question 99741: Factor completely using integer coefficients:
e) x^4+8x^3-2x^2-16x
f) 2x^6-20x^4-16x^3+160x
j) x^4-x^3-x+1

Answer by Earlsdon(6294) About Me  (Show Source):
You can put this solution on YOUR website!
Factor completely using integer coefficients:
e) x%5E4%2B8x%5E3-2x%5E2-16x First, factor an x.
x%28x%5E3%2B8x%5E2-2x-16%29 Now factor the parentheses. I know that my factors here will be two binomials of the form:
%28x%5E2%2Bm%29%28x%2Bn%29 where m%2An+=+-16 and nx%5E2+=+8x%5E2 and mx+=+-2x, so m = -2 and n = 8. So the parentheses, when factored, look like:
%28x%5E2-2%29%28x%2B8%29, then the final answer is:
x%5E4%2B8x%5E3-2x%5E2-16x+=+x%28x%5E2-2%29%28x%2B8%29
f) 2x%5E6-20x%5E4-16x%5E3%2B160x First, factor 2x.
2x%28x%5E5-10x%5E3-8x%5E2%2B80%29 Now factor the parentheses. I know my factors here will be two binomials of the form:%28x%5E3%2Bm%29%28x%5E2%2Bn%29 where: m%2An+=+80nx%5E3+=+-10x%5E3 and mx%5E2+=+-8x%5E2 so m+=+-8 and n+=+-10, so the parentheses, when factored, look like:
x%5E3-8%29%28x%5E2-10%29, so far we have:
2x%5E6-20x%5E4-16x%5E3%2B160x+=+2x%28x%5E3-8%29%28x%5E2-10%29 but notice that x%5E3-8 is the difference of two cubes and this can be factored.
x%5E3-8+=+%28x%29%5E3-%282%29%5E3 The difference of two cubes is factored thus:
A%5E3-B%5E3+=+%28A-B%29%28A%5E2%2BAB%2BB%5E2%29 so, in this case:
x%5E3-8+=+%28x-2%29%28x%5E2%2B2x%2B4%29 putting it all together, we have:
2x%5E6-20x%5E4-16x%5E3%2B160x+=+2x%28x-2%29%28x%5E2%2B2x%2B4%29%28x%5E2-10%29
j) x%5E4-x%5E3-x%2B1 Factor as:
%28x%5E3-1%29%28x-1%29 But again, we have the difference of two cubes:%28x%29%5E3-%281%29%5E3, so...
x%5E4-x%5E3-x%2B1+=+%28x-1%29%28x%5E2%2Bx%2B1%29%28x-1%29 or:
x%5E4-x%5E3-x%2B1+=+%28x-1%29%5E2%28x%5E2%2Bx%2B1%29