Polynomials in ℂ practice questions — SAT Math

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

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

Set the divisor to zero: $$2x + 1 = 0 \implies x = -\tfrac{1}{2}.$$

By the remainder theorem the remainder equals $f\!\left(-\tfrac{1}{2}\right)$: $$f\!\left(-\tfrac{1}{2}\right) = 4\left(-\tfrac{1}{8}\right) + 2\left(\tfrac{1}{4}\right) + 6\left(-\tfrac{1}{2}\right) + 5.$$

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

So the remainder is $2$.

$x_1 x_2 x_3 = -\dfrac{-4}{2} = 2$.

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

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

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

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

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, $$r_3 = \frac{-3}{-6} = \frac{1}{2}.$$

Here $b = -7$, $c = 12$.

By Vieta, sum of roots $= -b = -(-7) = 7$.

(Check by factoring: $(x-3)(x-4) = 0$, so roots are $3$ and $4$, summing to $7$.)