Remainder theorem tells about the remainder of division of a polynomial with x - a. The roots of given function are: 1, -1, -3
If there is a polynomial p(x), and a constant number 'a', then
[tex]\dfrac{p(x)}{(x-a)} = g(x) + p(a)[/tex]
where g(x) is a factor of p(x)
For the given case, we've got:
[tex]f(x) = x^3 + 3x^2 -x - 3[/tex]
and
f(1) = 0
Thus, for a = 1, from remainder theorem, we get:
[tex]\dfrac{p(x)}{(x-a)} = g(x) + p(a)\\\\\dfrac{f(x)}{(x-1)} = g(x) + f(1)\\\\\dfrac{x^3 + 3x^2 - x - 3}{(x-1)} = g(x) + 0\\\\ \dfrac{x^2(x+3) -1(x+3)}{x-1} = \dfrac{(x^2-1)(x+3)}{(x-1)} = g(x)\\\\(x+1)(x+3) = g(x)[/tex]
Thus, the other factor of f(x) is g(x) = (x+1)(x+3)
And we have:
[tex]\dfrac{f(x)}{x-1} = g(x)\\\\f(x) = (x+1)(x+3)(x-1)[/tex]
Putting it equal to 0, we get:
[tex]f(x) = 0\\(x+1)(x+3)(x-1) = 0\\x = 1, -1, -3[/tex]
Thus,
The roots of given function are: 1, -1, -3
Learn more about polynomial remainder theorem here:
https://brainly.com/question/10488065
Answer:
A
Step-by-step explanation:
got it right on Edge 1943 whilst fighting the German Forces in Northern Italy while my friends were on the beaches of Normandy.
