document.write( "Question 583689: Can you please help me solve this? I am supposed to solve and state whether the binomial is a factor of the polynomial.\r
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document.write( "I used synthetic division to drop the coefficients and got: \r
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document.write( "2] 16 -32 0 0 -81 162
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document.write( " 0 32 0 0 0 -162
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document.write( " 16 0 0 0 0 -81 0
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document.write( "(The zeroes in the top row are because there are no x-cubed and x-squared variables, therefore they're respresented by a zero, right?)\r
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document.write( "Multiple choice allowed me the answers of:
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document.write( "A - 1348; no
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document.write( "B - 0; yes
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document.write( "C - 700; yes
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document.write( "D - 0; no\r
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document.write( " I am pretty sure the answer is C because my remainder is zero, and x-2 is a factor of the original equation. If this is not right, can you please show me what I've done wrong?
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document.write( "(Also, I apologize if the synthetic division does not line up properly when submitted and if it makes it harder for you to read.) \n" );
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Algebra.Com's Answer #372534 by richwmiller(17219) You can put this solution on YOUR website! If your remainder is zero then why isn't it B. \n" ); document.write( "(16x^5-32x^4-81x+162)/(x-2) does equal 16x^4-81 which can be factored to \n" ); document.write( "(-3+2x)*(3+2x)*(9+4x^2) \n" ); document.write( " |