How do you factor $x^{3} + x^{2} - 7 x - 3$ $x^3+x^2-7x-3$ ?

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How do you factor $x^{3} + x^{2} - 7 x - 3$ $x^3+x^2-7x-3$ ?

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Answer 1 · 2016-08-18T19:50:05

$x^{3} + x^{2} - 7 x - 3 = (x + 3) (x - 1 - \sqrt{2}) (x - 1 + \sqrt{2})$ $x^3+x^2-7x-3 = (x+3)(x-1-sqrt(2))(x-1+sqrt(2))$

Explanation:

$f (x) = x^{3} + x^{2} - 7 x - 3$ $f(x) = x^3+x^2-7x-3$

By the rational roots theorem, any rational zeros of $f (x)$ $f(x)$ are expressible in the form $\frac{p}{q}$ $p/q$ for integers $p, q$ $p, q$ with $p$ $p$ a divisor of the constant term $- 3$ $-3$ and $q$ $q$ a divisor of the coefficient $1$ $1$ of the leading term.

That means tha the only possible rational zeros are:

$\pm 1, \pm 3$ $+-1, +-3$

We find:

$f (- 3) = - 27 + 9 + 21 - 3 = 0$ $f(-3) = -27+9+21-3 = 0$

So $x = - 3$ $x=-3$ is a zero and $(x + 3)$ $(x+3)$ a factor:

$x^{3} + x^{2} - 7 x - 3 = (x + 3) (x^{2} - 2 x - 1)$ $x^3+x^2-7x-3 = (x+3)(x^2-2x-1)$

We can factor the remaining quadratic by completing the square and using the difference of squares identity:

$a^{2} - b^{2} = (a - b) (a + b)$ $a^2-b^2=(a-b)(a+b)$

with $a = (x - 1)$ $a=(x-1)$ and $b = \sqrt{2}$ $b=sqrt(2)$ as follows:

$x^{2} - 2 x - 1$ $x^2-2x-1$

$= x^{2} - 2 x + 1 - 2$ $=x^2-2x+1-2$

$= {(x - 1)}^{2} - {(\sqrt{2})}^{2}$ $=(x-1)^2-(sqrt(2))^2$

$= (x - 1 - \sqrt{2}) (x - 1 + \sqrt{2})$ $=(x-1-sqrt(2))(x-1+sqrt(2))$

Putting it all together:

$x^{3} + x^{2} - 7 x - 3 = (x + 3) (x - 1 - \sqrt{2}) (x - 1 + \sqrt{2})$ $x^3+x^2-7x-3 = (x+3)(x-1-sqrt(2))(x-1+sqrt(2))$

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How do you factor $x^{3} + x^{2} - 7 x - 3$ $x^3+x^2-7x-3$ ?

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How do you factor $x^{3} + x^{2} - 7 x - 3$ $x^3+x^2-7x-3$ ?