document.write( "Question 99741: Factor completely using integer coefficients:
\n" ); document.write( "e) x^4+8x^3-2x^2-16x\r
\n" ); document.write( "\n" ); document.write( "f) 2x^6-20x^4-16x^3+160x\r
\n" ); document.write( "\n" ); document.write( "j) x^4-x^3-x+1
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Algebra.Com's Answer #72636 by Earlsdon(6294)\"\" \"About 
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Factor completely using integer coefficients:
\n" ); document.write( "e) \"x%5E4%2B8x%5E3-2x%5E2-16x\" First, factor an x.
\n" ); document.write( "\"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:
\n" ); document.write( "\"%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:
\n" ); document.write( "\"%28x%5E2-2%29%28x%2B8%29\", then the final answer is:
\n" ); document.write( "\"x%5E4%2B8x%5E3-2x%5E2-16x+=+x%28x%5E2-2%29%28x%2B8%29\"\r
\n" ); document.write( "\n" ); document.write( "f) \"2x%5E6-20x%5E4-16x%5E3%2B160x\" First, factor 2x.
\n" ); document.write( "\"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+=+80\"\"nx%5E3+=+-10x%5E3\" and \"mx%5E2+=+-8x%5E2\" so \"m+=+-8\" and \"n+=+-10\", so the parentheses, when factored, look like:
\n" ); document.write( "\"x%5E3-8%29%28x%5E2-10%29\", so far we have:
\n" ); document.write( "\"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.
\n" ); document.write( "\"x%5E3-8+=+%28x%29%5E3-%282%29%5E3\" The difference of two cubes is factored thus:
\n" ); document.write( "\"A%5E3-B%5E3+=+%28A-B%29%28A%5E2%2BAB%2BB%5E2%29\" so, in this case:
\n" ); document.write( "\"x%5E3-8+=+%28x-2%29%28x%5E2%2B2x%2B4%29\" putting it all together, we have:
\n" ); document.write( "\"2x%5E6-20x%5E4-16x%5E3%2B160x+=+2x%28x-2%29%28x%5E2%2B2x%2B4%29%28x%5E2-10%29\"\r
\n" ); document.write( "\n" ); document.write( "j) \"x%5E4-x%5E3-x%2B1\" Factor as:
\n" ); document.write( "\"%28x%5E3-1%29%28x-1%29\" But again, we have the difference of two cubes:\"%28x%29%5E3-%281%29%5E3\", so...
\n" ); document.write( "\"x%5E4-x%5E3-x%2B1+=+%28x-1%29%28x%5E2%2Bx%2B1%29%28x-1%29\" or:
\n" ); document.write( "\"x%5E4-x%5E3-x%2B1+=+%28x-1%29%5E2%28x%5E2%2Bx%2B1%29\"
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