Question

Sekkrit help!!!!! If (x+1) is the factor of polynomial p(x) = ax²+x+1, then find a.

Answer 1

Answer:

The value of a is 0.

Step-by-step explanation:

Given that (x+1) is a factor to a function, it means that when x = -1 is substitute into the function, you will get a 0 value. So you have to substitute the value of x into the function and make it 0, to find a :

$p(x) = a {x}^{2} + x + 1$

$let \: p( - 1) = 0 \\ let \: x = -1$

$p( - 1) = a {( - 1)}^{2} + ( - 1) + 1$

$0 = a - 1 + 1$

$a = 0$

Answer 2

Answer:

a=0

Solution,

To find a,

We should know that,

Factor of polynomial gives root of polynomial like:x-a if a factor of p(X) then p(a)=0 at X=a

So,

X+1=0

X=0-1

X=-1

put x=-1 into p(X) it gives zero.

$p( - 1) = 0 \\ a {( - 1)}^{2} + ( - 1) + 1 = 0 \\ a(1) - 1 + 1 = 0 \\ a = 0$

hope this helps....

Good luck on your assignment....