Question

Using long division method shown that, p(x) = x²+1 is divisible by q(x) = x+1. Verify your result using factor theorum.​

Answer 1

Answer:

No, $x^{2}$ + 1 is not divisible by x + 1

Step-by-step explanation:

Factor theorem says: if f(x) is a polynomial of degree n ≥ 1 and ‘k’ is any real number, then, (x-k) is a factor of f(x), if f(k) = 0

So, here p(x) = $x^{2}$ + 1
And q(x) =  x + 1

Now, x + 1 = 0
    ⇒ x = -1

Now, put -1 in p(x) and if it becomes 0 then q(x) is a factor according to the factor theorem.

So, p(x) = $x^{2}$ + 1
   ⇒ p(-1) = $(-1)^{2}$ + 1
   ⇒ p(-1) = 1 + 1 = 2
∴ p (-1) $\neq$ 0
     Hence p(x) isn't divisible by q(x).

Also, you can check using the long division method.