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

Question 1

For which value of $m \in \mathbb{R}$ is $X = 2$ a root of $P(X) = X^3 - mX + 4$?

Accepted Answer

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

Answer

$-6$

Answer

$-4$

Answer

$6$

Answer

$8$

Question 2

The number of real solutions of $X^4 - 5X^2 + 4 = 0$ is:

Accepted Answer

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.

Answer

$1$

Answer

$2$

Answer

$3$

Answer

$4$

Question 3

The polynomial $P(X) = X^3 - X^2 - 4X + 4$ factors as:

Accepted Answer

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

Answer

$(X - 1)(X - 2)(X + 2)$

Answer

$(X + 1)(X - 2)(X + 2)$

Answer

$(X - 1)^2(X + 4)$

Answer

$(X - 1)(X + 2)^2$

Question 4

For the polynomial $P(X) = 2X^3 - 6X^2 + X - 4$ with roots $x_1, x_2, x_3$, the product $x_1 x_2 x_3$ equals:

Accepted Answer

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

Answer

$-4$

Answer

$-2$

Answer

$2$

Answer

$4$

Question 5

For $P(X) = X^3 - 6X^2 + 11X - 6$ with roots $x_1, x_2, x_3$, the value of $x_1 + x_2 + x_3$ is:

Accepted Answer

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 $1, 2, 3$.)

Answer

$-11$

Answer

$-6$

Answer

$6$

Answer

$11$

Question 6

Dividing $x^{2} - 5x + 6$ by $(x - 2)$ gives the quotient:

Accepted Answer

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

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

Answer

$x - 6$

Answer

$x + 3$

Answer

$x - 3$

Answer

$x - 2$

Question 7

The remainder of the division of $P(X) = X^3 + 2X^2 - X + 5$ by $X - 1$ is:

Accepted Answer

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

Answer

$0$

Answer

$5$

Answer

$7$

Answer

$9$

Question 8

Expand $(x + 2)(x - 5)$.

Accepted Answer

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

Answer

$x^{2} - 3x - 10$

Answer

$x^{2} - 3x - 10$

Answer

$x^{2} + 3x + 10$

Answer

$x^{2} - 7x - 10$

Question 9

The polynomial is defined by
$$f(x) = 4x^3 + 2x^2 + 6x + 5.$$
Find the remainder when $f(x)$ is divided by $(2x + 1)$.

Accepted Answer

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

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

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

So the remainder is $2$.

Answer

$-3$

Answer

$2$

Answer

$9$

Answer

$17$

Question 10

The quotient when $P(X) = X^3 - 4X^2 + 5X - 2$ is divided by $X - 2$ has degree:

Accepted Answer

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

Answer

$1$

Answer

$2$

Answer

$3$

Answer

$4$

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

Problems & worked solutions

Problem 1 — Polynomial functions, roots and factors

Problem 2 — Polynomial functions, roots and factors

Problem 3 — Polynomial functions, roots and factors

Problem 4 — Polynomial functions, roots and factors

Problem 5 — Polynomial functions, roots and factors

Problem 6 — Polynomial functions, roots and factors

Problem 7 — Polynomial functions, roots and factors

Problem 8 — Polynomial functions, roots and factors

Problem 9 — Polynomial functions, roots and factors

Problem 10 — Polynomial functions, roots and factors