Polynomials in ℂ practice questions — A-Level Maths

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.

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

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

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

For $p(x)=2x^3-3x^2-11x+6$, the product of the three roots is $$r_1{r}_2{r}_3=-\frac{d}{a}=-\frac{6}{2}=-3.$$ Substituting the two known roots $r_1=3$ and $r_2=-2$, $$3\cdot (-2)\cdot r_3=-3.$$ Solving for the third root, $$r_3=\frac{-3}{-6}=\frac{1}{2}\mathrm{.}$$

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

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

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

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 $1,2,3$.)

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

$x^2-9=x^2-3^2=(x-3)(x+3)$ — the difference-of-squares identity.