document.write( "Question 100274: If x+p is a factor of ax*+bx+3, show that p(b-ap)=3. \n" ); document.write( "
Algebra.Com's Answer #73019 by jim_thompson5910(35256)\"\" \"About 
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Before I start, if \"ax%5E2%2Bbx%2B3=0\", then \"ax%5E2%2Bbx=-3\". So keep this fact in mind\r
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\n" ); document.write( "\n" ); document.write( "\"p%28b-ap%29\" Start with the left side of the second expression\r
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\n" ); document.write( "\n" ); document.write( "=\"bp-ap%5E2\" Distribute\r
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\n" ); document.write( "\n" ); document.write( "Since we're assuming x+p is a factor, this means \"x=-p\" which means \"p=-x\"\r
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\n" ); document.write( "\n" ); document.write( "=\"b%28-x%29-a%28-x%29%5E2\" Plug in \"p=-x\"\r
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\n" ); document.write( "\n" ); document.write( "=\"-bx-ax%5E2\" Simplify\r
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\n" ); document.write( "\n" ); document.write( "=\"-ax%5E2-bx\" Rearrange the terms\r
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\n" ); document.write( "\n" ); document.write( "=\"-%28ax%5E2%2Bbx%29\" Factor out a negative 1\r
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\n" ); document.write( "\n" ); document.write( "Remember \"ax%5E2%2Bbx=-3\", so let's substitute that in\r
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\n" ); document.write( "\n" ); document.write( "=\"-%28-3%29\" Replace \"ax%5E2%2Bbx\" with -3\r
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\n" ); document.write( "\n" ); document.write( "=\"3\" Negate -3 to get 3\r
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\n" ); document.write( "\n" ); document.write( "So we've algebraically manipulated \"p%28b-ap%29\" to get 3. So we've just shown that \"p%28b-ap%29\" is equal to 3\r
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\n" ); document.write( "\n" ); document.write( "So that means we've proven that \"p%28b-ap%29=3\" is true.
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