# How do you show that x+1 is a factor of x^4+2x^3-2x^2-6x-3?

Jan 7, 2017

See explanation.

#### Explanation:

If ${P}_{n} \left(x\right) - {a}_{0} {x}^{n} + {a}_{1} {x}^{n - 1} + \ldots + {a}_{n - 1} x + {a}_{n}$;

the sum of the coefficients

${P}_{n} \left(1\right) = {a}_{0} + {a}_{1} + {a}_{2} + \ldots + {a}_{n - 1} + {a}_{n} = 0$

shows that 1 is a zero of P_n. and so,

(x-1) is a factor of ${P}_{n} \left(x\right)$, Likewise, if

P_(n-1) =(-1)^n(a_0-a_1+a_2-...+(-1)^a_n=0, then

$- 1$ is a zero of P_n(x), and so,

(x-(-1)=(x+1) is a factor.

Here, ${P}_{4} \left(- 1\right) = 1 - 2 - 2 + 6 - 3 = 2$, and so, ( x+1) is a factor.

Jan 7, 2017

By using the factor theorem

$\left(x - a\right)$ is a factor of $P \left(x\right) \implies P \left(a\right) = 0$

#### Explanation:

proof

Let $P \left(x\right)$ be divided by $\left(x - a\right)$ giving quotient $Q \left(x\right)$and remainder $R$

then

$P \left(x\right) = \left(x - a\right) Q \left(x\right) + R$

$P \left(a\right) = \cancel{\left(a - a\right) Q \left(x\right)} + R$

$R = 0 \implies \left(x - a\right)$ is a factor of $P \left(x\right)$

in this case

to show $\left(x + 1\right)$is a factor of

$P \left(x\right) = {x}^{4} + 2 {x}^{3} - 2 {x}^{2} - 6 x - 3$

$a = - 1$

$P \left(- 1\right) = {\left(- 1\right)}^{4} + 2 \times {\left(- 1\right)}^{3} - 2 \times {\left(- 1\right)}^{2} - 6 \times \left(- 1\right) - 3$

$P \left(- 1\right) = 1 - 2 - 2 + 6 - 3 = - 3 + 6 - 3 = 0$

$P \left(- 1\right) = 0 \implies \left(x + 1\right)$ is a a factor $P \left(x\right)$