Call the polynomial p(x)$p(x)$. We use Horner’s method to calculate the value of p(x)$p(x)$ in the point x=2$x=2$, and we get the quotient for free. You can, of course, also calculate p(2)$p(2)$ with your calculator, but then you still have to use Horner or Euclidean Division if you need the quotient.

The starting scheme looks like this

where the coefficients of p(x)$p(x)$ are in the top row. We now drop the first coefficient (i.e. 2$2$), multiply it by the 2$2$ on the left, and write the result in the second row, under the second coefficient (i.e. under −4$-4$):

We add the −4$-4$ and 4$4$ together, write the result −4+4=0$-4+4=0$ at the bottom and multiply 0$0$ by the left 2$2$ and write the result 0$0$ again in the second row, under the 5$5$, and then add 0$0$ to that 5$5$:

We again multiply the bottom 5$5$ with the left 2$2$, write the product 10$10$ under −7$-7$, add up and get the number 3$3$ at the bottom right:

The scheme is now complete, and we can prove that the rightmost number at the bottom is always the numerical value of the polynomial in the point 2$2$, and that the other numbers in the bottom row are always the coefficients of the quotient of p(x)$p(x)$ divided by x−2$x-2$.

So the quotient is 2x2+0x+5$2x^2+0x+5$ and the remainder is 3$3$, or

Thus: 2x3+x2−5x+2=(2x2+7x+16)(x−3)+50$2x^3+x^2-5x+2=(2x^2+7x+16)(x-3)+50$.

Note that the remainder 50$50$ thus also represents the numerical value of the dividend for x=3$x=3$, because

Notice already that this is the same task as ’factor this polynomial’.

For candidate divisors of p(x)$p(x)$ of the form x−a$x-a$, with a$a$ an integer, a$a$ must be a divisor of 24$24$. A possible way to solve this exercise is to check for all integer divisors of 24 whether their numerical value is zero. If you start with the smallest one, you are immediately in luck, because p(1)=1−15−10+24=0$p(1)=1-15-10+24=0$, and so x−1$x-1$ is a divisor of p(x)$p(x)$. We can now search for the numerical values of the divisors −1,2,−2,3,−3,4,−4,6,−6,12,−12,24$-1,2,-2,3,-3,4,-4,6,-6,12,-12,24$ and −24$-24$, but it is easier to calculate with Horner that

and therefore x4−15x2−10x+24=(x−1)(x3+x2−14x−24)$x^4-15x^2-10x+24=(x-1)(x^3+x^2-14x-24)$. We repeat the question for the polynomial x3+x2−14x−24$x^3+x^2-14x-24$. The final result will be p(x)=(x−1)(x+2)(x+3)(x−4)$p(x)=(x-1)(x+2)(x+3)(x-4)$. This is very clearly and fully explained in the following video, in which Horner’s own rule is also explained at the beginning.

Since p(−3)=0$p(-3)=0$ is x−(−3)=x+3$x-(-3)=x+3$ a divisor of x4−10x2+9$x^4-10x^2+9$.

We calculate the quotient with Horner’s method.

We now look further for the zero points of x3−3x2−x+3$x^3-3x^2-x+3$, and find (after some more calculations) that x3−3x2−x+3=(x−1)(x+1)(x−3)(x+3)$x^3-3x^2-x+3=(x-1)(x+1)(x-3)(x+3)$ .

Note: in this case the polynomial can be factorised (much) more easily by finding all zero points.