document.write( "Question 799653: ax^3+bx^2+x-6 has factor (x+2) and (x-2) then in both cases remainder is 4 then find the value of a and b? thanks if anybody can solve it for me \n" ); document.write( "

Algebra.Com's Answer #482748 by josgarithmetic(39623) $\"\"$ $\"About$

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Many details to the process, but you would want to do synthetic division using \"divisors\" of x+2, and of x-2, each separately. The roots correspondingly being checked are -2 and +2. \r
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\n" ); document.write( "\n" ); document.write( "The dividend would be one of the factors of $\"ax%5E3%2Bbx%5E2%2Bx%2B6\"$ , once a has been factored. This dividend is $\"x%5E3%2B%28b%2Fa%29x%5E2%2Bx%2Fa%2B6%2Fa\"$ . During the division, the intermediary expressions become more complicated...\r
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\n" ); document.write( "\n" ); document.write( "The root, -2 gives remainder $\"%284%2B4b-8a%29%2Fa=4\"$ , equated according to the description of the problem; and the root +2 gives remainder $\"%288%2B4b%2B8a%29%2Fa=4\"$ .
\n" ); document.write( "These seem to be two equations in the unknowns, a and b. \r
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\n" ); document.write( "\n" ); document.write( "Finishing the solution for a and b from here should be no trouble, seems to appear straightforward. \n" ); document.write( "

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