Question

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

Answer 1

See explanation.

Explanation:

If `P_n(x) - a_0x^n+a_1x^(n-1)+...+a_(n-1)x+a_n`;

the sum of the coefficients

`P_n(1) = a_0+a_1+a_2+...+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(x)`, 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(-1)=1-2-2+6-3=2`, and so, ( x+1) is a factor.

Answer 2

By using the factor theorem

`(x-a)` is a factor of `P(x)=> P(a)=0`

Explanation:

proof

Let `P(x)` be divided by `(x-a)` giving quotient `Q(x)`and remainder `R`

then

`P(x)=(x-a)Q(x)+R`

`P(a)=cancel((a-a)Q(x))+R`

`R=0=>(x-a)` is a factor of `P(x)`

in this case

to show `(x+1)`is a factor of

`P(x)=x^4+2x^3-2x^2-6x-3`

`a=-1`

`P(-1)=(-1)^4+2xx(-1)^3-2xx(-1)^2-6xx(-1)-3`

`P(-1)=1-2-2+6-3=-3+6-3=0`

`P(-1)=0=>(x+1)` is a a factor `P(x)`