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March 14, 2026Mathematics

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2 parts

(2x-1)(x+3)

2x + 1 + \dfrac{-2x + 3}{x^{2} - 1}

10 steps

Factor $2x^2 + 5x - 3$ .
Find two numbers that multiply to $a \cdot c = 2 \cdot (-3) = -6$ and add to $b = 5$ : 6 and $-1$ .
$2x^2 + 6x - x - 3$
Group terms.
$2x(x + 3) - 1(x + 3)$
Factor out common binomial.
$(2x - 1)(x + 3)$
(2x-1)(x+3)

Simplify $\dfrac{2x^3 + x^2 - 4x + 2}{x^2 - 1}$ using polynomial division.
Divide $2x^3$ by $x^2$ :
$2x$ .
Multiply $2x$ by $x^2 - 1$
$2x(x^2 - 1) = 2x^3 - 2x$
Subtract
$\begin{align*} &(2x^3 + x^2 - 4x + 2) - (2x^3 - 2x)\\ &= x^2 - 2x + 2 \end{align*}$
Divide $x^2$ by $x^2$
Multiply 1 by $x^2 - 1$
$x^2 - 1$
Subtract
$\begin{align*} &(x^2 - 2x + 2) - (x^2 - 1)\\ &= -2x + 3 \end{align*}$
Quotient is $2x + 1$ , remainder $-2x + 3$ .
$\dfrac{2x^3 + x^2 - 4x + 2}{x^2 - 1} = 2x + 1 + \dfrac{-2x + 3}{x^2 - 1}$
Final answer:
2x + 1 + \dfrac{-2x + 3}{x^{2} - 1}

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