Originally Posted by
Jen1603 Use the remainder theorem to factorise :
1) $\displaystyle f(x)=x^3 +3x^2 - 4$
The rational roots theorem tells us if this has rational roots, they will be
$\displaystyle \pm 1, \pm 2, \pm4$
Test each one using the remainder theorem to see which one(s) satisfy $\displaystyle f(x)=0$
I found $\displaystyle f(-2)=0$, so $\displaystyle x+2$ is a factor.
To find the remaining factors, you could test the rest of the possible rational roots using the remainder theorem or use synthetic division with divisor -2to reduce the polynomal to a quadratic that you can factor.
Code:
-2 | 1 3 0 -4
-2 -2 4
------------
1 1 -2 0
The quadratic remaining after factoring $\displaystyle x+2$ is $\displaystyle x^2+x-2$
That factors into $\displaystyle (x+2)(x-1)$
$\displaystyle \boxed{f(x)=(x+2)(x+2)(x-1)}$
Originally Posted by
Jen1603 2)$\displaystyle f(x)= x^4 + 3x^3 + 5x^2 + 9x + 6$
Test these factors $\displaystyle \pm 1, \pm 2, \pm 3, \pm 6$
There are no sign changes in this polynomial, so no need to check for positive zero because there aren't any (Descartes Rule of Signs)