Find the quotient and the remainder $(4 x^{3} + 6 x^{2} - 23 x - 15) \div (3 + x)$

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Solution

Write the given polynomials in ascending or decending order.
$(4 x^{3} + 6 x^{2} - 23 x - 15) \div (3 + x)$
(i) $\frac{4 x^{3}}{x} = 4 x^{2}$
(ii) $4 x^{2} (x + 3) = 4 x^{3} + 12 x^{2}$
(iii) $\frac{- 6 x^{2}}{x} = 6 x$
(iv) $- 6 x (x + 3) = - 6 x^{2} - 18 x$
(v) $\frac{- 5 x}{x} = - 5$
(vi) $- 5 (x + 3) = - 5 x - 15$
$∴$ Quotient $= 4 x^{2} - 6 x - 5$
Remainder =0

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