How do you express $\frac{2 x^{3} - x^{2} + x + 5}{x^{2} + 3 x + 2}$ $(2x^3-x^2+x+5)/(x^2+3x+2)$ in partial fractions?

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How do you express $\frac{2 x^{3} - x^{2} + x + 5}{x^{2} + 3 x + 2}$ $(2x^3-x^2+x+5)/(x^2+3x+2)$ in partial fractions?

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Answer 1 · 2016-12-20T19:12:18

The answer is $= 2 x - 7 + \frac{1}{x + 1} + \frac{17}{x + 2}$ $=2x-7+1/(x+1)+17/(x+2)$

Explanation:

As the degree of the numerator is $>$ $>$ than the degree of the denominator, we do a long division

$a a a a$ $color(white)(aaaa)$ $2 x^{3} - x^{2} + x + 5$ $2x^3-x^2+x+5$ $a a a a$ $color(white)(aaaa)$ $∣$ $∣$ $x^{2} + 3 x + 2$ $x^2+3x+2$

$a a a a$ $color(white)(aaaa)$ $2 x^{3} + 6 x^{2} + 4 x$ $2x^3+6x^2+4x$ $a a a a a a$ $color(white)(aaaaaa)$ $∣$ $∣$ $2 x - 7$ $2x-7$

$a a a a a$ $color(white)(aaaaa)$ $0 - 7 x^{2} - 3 x + 5$ $0-7x^2-3x+5$

$a a a a a a a$ $color(white)(aaaaaaa)$ $- 7 x^{2} - 21 x - 14$ $-7x^2-21x-14$

$a a a a a a a a a a a a a a$ $color(white)(aaaaaaaaaaaaaa)$ $18 x + 19$ $18x+19$

Therefore,

$\frac{2 x^{3} - x^{2} + x + 5}{x^{2} + 3 x + 2} = 2 x - 7 + \frac{18 x + 19}{x^{2} + 3 x + 2}$ $(2x^3-x^2+x+5)/(x^2+3x+2)=2x-7+(18x+19)/(x^2+3x+2)$

We can now do the decomposition into partial fractions

$\frac{18 x + 19}{x^{2} + 3 x + 2} = \frac{18 x + 19}{(x + 1) (x + 2)} = \frac{A}{x + 1} + \frac{B}{x + 2}$ $(18x+19)/(x^2+3x+2)=(18x+19)/((x+1)(x+2))=A/(x+1)+B/(x+2)$

$= \frac{A (x + 2) + B (x + 1)}{(x + 1) (x + 2)}$ $=(A(x+2)+B(x+1))/((x+1)(x+2))$

So,

$18 x + 19 = A (x + 2) + B (x + 1)$ $18x+19=A(x+2)+B(x+1)$

Let $x = - 1$ $x=-1$ , $\Rightarrow$ $=>$ , $1 = A$ $1=A$

Let $x = - 2$ $x=-2$ , $\Rightarrow,$ $=>,$ -17=-B#

$\frac{2 x^{3} - x^{2} + x + 5}{x^{2} + 3 x + 2} = 2 x - 7 + \frac{1}{x + 1} + \frac{17}{x + 2}$ $(2x^3-x^2+x+5)/(x^2+3x+2)=2x-7+1/(x+1)+17/(x+2)$

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How do you express $\frac{2 x^{3} - x^{2} + x + 5}{x^{2} + 3 x + 2}$ $(2x^3-x^2+x+5)/(x^2+3x+2)$ in partial fractions?

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Explanation:

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How do you express $\frac{2 x^{3} - x^{2} + x + 5}{x^{2} + 3 x + 2}$ $(2x^3-x^2+x+5)/(x^2+3x+2)$ in partial fractions?