How do you use polynomial long division to divide $(4 x^{2} - x - 23) \div (x^{2} - 1)$ $(4x^2-x-23)div(x^2-1)$ and write the polynomial in the form $p (x) = d (x) q (x) + r (x)$ $p(x)=d(x)q(x)+r(x)$ ?

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Answer 1 · 2017-11-29T07:51:27

$p (x) = d (x^{2} - 1) q (4 x - 1) + r (- 23)$ $p(x)=d(x^2-1)q(4x-1)+r(-23)$

$x^{2} - 1 \div 4 x^{2} - x - 23$ $x^2-1-:4x^2-x-23$

do not forget placeholders!

$x^{2} + 0 x - 1 \div 4 x^{2} - x - 23$ $x^2+0x-1-:4x^2-x-23$
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dividend: $(4 x^{2} - x - 23)$ $(4x^2-x-23)$
divisor: $(x^{2} - 1)$ $(x^2-1)$
quotient: $(4 x - 1)$ $(4x-1)$
remainder: $- 23$ $-23$

How do you use polynomial long division to divide (4x^2-x-23)div(x^2-1)(4x2−x−23)÷(x2−1)(4x^2-x-23)div(x^2-1) and write the polynomial in the form p(x)=d(x)q(x)+r(x)p(x)=d(x)q(x)+r(x)p(x)=d(x)q(x)+r(x)?