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

2 Answers
Jan 7, 2017

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.

Jan 7, 2017

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)