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 $\dfrac{x^{2} - 5x + 6}{x - 2} = x - 3$, 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, $$r_3 = \frac{-3}{-6} = \frac{1}{2}.$$

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

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

$P(1) = 0$, 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 $1, 2, 3$.)

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

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