Polynomial functions, roots and factors practice questions — IB Maths AA

$P(2)=8-2m+4=12-2m=0\Rightarrow m=6$.

With $Y=X^2$: $Y^2-5Y+4=0\Rightarrow Y\in \{1,4\}$. Then $X^2=1\Rightarrow X=\pm 1$ and $X^2=4\Rightarrow X=\pm 2$. Four real solutions.

$P(1)=0$, so $X-1$ divides $P$. By Horner: $P(X)=(X-1)(X^2-4)=(X-1)(X-2)(X+2)$.

$x_1{x}_2{x}_3=-{\displaystyle \frac{-4}{2}}=2$.

The leading coefficient is $1$ and the coefficient of $X^2$ is $-6$, so $x_1+x_2+x_3=-(-6)=6$. (The roots are in fact $1,2,3$.)

$x^2-5x+6=(x-2)(x-3)$.

So ${\displaystyle \frac{x^2-5x+6}{x-2}}=x-3$, with no remainder.

$P(1)=1+2-1+5=7$.

$(x+2)(x-5)=x^2-5x+2x-10=x^2-3x-10$.

Set the divisor to zero: $$2x+1=0⟹x=-\frac{1}{2}\mathrm{.}$$

By the remainder theorem the remainder equals $f\text{\hspace{0.17em}⁣}(-\frac{1}{2})$: $$f\text{\hspace{0.17em}⁣}(-\frac{1}{2})=4(-\frac{1}{8})+2\left(\frac{1}{4}\right)+6(-\frac{1}{2})+5.$$

Evaluate term by term: $$-\frac{1}{2}+\frac{1}{2}-3+5=2.$$

So the remainder is $2$.

$\mathrm{deg}(Q)=3-1=2$.