Polynomials -- Diagnostic Tests

Tests edge cases, boundary conditions, and common misconceptions for polynomials.

Question:

Determine whether $x + 2$ is a factor of $P(x) = x^3 + 3x^2 - 4x - 8$.

Solution:

By the factor theorem, $x + 2$ is a factor if and only if $P(-2) = 0$.

$P(-2) = (-2)^3 + 3(-2)^2 - 4(-2) - 8 = -8 + 12 + 8 - 8 = 4$

Since $P(-2) = 4 \neq 0$, $x + 2$ is not a factor.

A common mistake is testing $P(2)$ instead of $P(-2)$. The factor theorem states that $(x - a)$ is a factor if $P(a) = 0$. Here the factor is $x + 2 = x - (-2)$ So we test $a = -2$.

Question:

When $P(x) = 2x^3 + ax^2 - 5x + 3$ is divided by $x - 2$The remainder is $11$. When $P(x)$ is divided by $x + 1$The remainder is $-4$. Find $a$ and $b$ (if $P(x)$ also has a constant term correction $b$ replacing $3$), or find $a$.

Solution:

By the remainder theorem:

$P(2) = 2(8) + 4a - 10 + 3 = 16 + 4a - 7 = 4a + 9 = 11 \implies 4a = 2 \implies a = \dfrac{1}{2}$.

Verify: $P(-1) = 2(-1) + \dfrac{1}{2}(1) - 5(-1) + 3 = -2 + \dfrac{1}{2} + 5 + 3 = 6.5 = -4$?

Wait: $P(-1) = 2(-1)^3 + \dfrac{1}{2}(-1)^2 - 5(-1) + 3 = -2 + \dfrac{1}{2} + 5 + 3 = \dfrac{13}{2}$.

But we need $P(-1) = -4$. There is no single value of $a$ satisfying both conditions simultaneously with the constant term fixed at $3$. The problem as stated is inconsistent. This highlights the importance of verifying all conditions.

If instead the constant term is also a variable $b$:

$P(2) = 4a + b + 9 = 11 \implies 4a + b = 2$.

$P(-1) = -2 + a + 5 + b = a + b + 3 = -4 \implies a + b = -7$.

Subtracting: $3a = 9 \implies a = 3$, $b = -10$.

Question:

Divide $2x^3 - 5x^2 + x - 6$ by $x^2 - 2x + 3$.

Solution:

Quotient: $2x - 1$Remainder: $-7x - 3$.

Verify: $(2x - 1)(x^2 - 2x + 3) + (-7x - 3)$

$= 2x^3 - 4x^2 + 6x - x^2 + 2x - 3 - 7x - 3$

$= 2x^3 - 5x^2 + x - 6$. Correct.

Question:

If $\alpha$, $\beta$, $\gamma$ are the roots of $x^3 - 5x^2 + 2x + 8 = 0$Find:

(a) $\alpha + \beta + \gamma$ (b) $\alpha\beta + \beta\gamma + \gamma\alpha$ (c) $\alpha\beta\gamma$

Solution:

By Vieta’s formulas for $x^3 + px^2 + qx + r = 0$:

$\alpha + \beta + \gamma = -p = 5$ $\alpha\beta + \beta\gamma + \gamma\alpha = q = 2$ $\alpha\beta\gamma = -r = -8$

Question:

$P(x) = x^3 + ax^2 + bx - 12$ has factors $(x - 1)$ and $(x + 4)$. Find $a$, $b$ And the remaining factor.

Solution:

Since $(x - 1)$ is a factor: $P(1) = 1 + a + b - 12 = 0 \implies a + b = 11$. … (1)

Since $(x + 4)$ is a factor: $P(-4) = -64 + 16a - 4b - 12 = 0 \implies 16a - 4b = 76 \implies 4a - b = 19$. … (2)

From (1) + (2): $5a = 30 \implies a = 6$. Then $b = 5$.

So $P(x) = x^3 + 6x^2 + 5x - 12$.

Dividing by $(x - 1)(x + 4) = x^2 + 3x - 4$:

$x^3 + 6x^2 + 5x - 12 = (x^2 + 3x - 4)(x + c)$

Expanding RHS: $x^3 + cx^2 + 3x^2 + 3cx - 4x - 4c = x^3 + (c+3)x^2 + (3c-4)x - 4c$.

Matching: $c + 3 = 6 \implies c = 3$.

The remaining factor is $(x + 3)$.

Tests synthesis of polynomials with other topics.

Question:

Let $P(x) = (x - 1)(x^2 - 4x + 3)$. Find the set of values of $x$ for which $P(x) \leq 0$.

Solution:

$P(x) = (x - 1)(x - 1)(x - 3) = (x - 1)^2(x - 3)$

Critical values: $x = 1$ (double root) and $x = 3$.

$P(x) \leq 0$ when $x \leq 3$ (including $x = 1$ and $x = 3$).

Therefore $x \in (-\infty,\; 3]$.

Question:

Let $f(x) = x^3 - 3x^2 - 4x + 12$. Given that $(x - 2)$ is a factor, find all $x$ for which $f(x) = 0$ And hence state the domain on which $f$ is one-to-one.

Solution:

$P(2) = 8 - 12 - 8 + 12 = 0$. Confirmed.

Divide $x^3 - 3x^2 - 4x + 12$ by $(x - 2)$:

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

Roots: $x = -2$, $x = 2$, $x = 3$.

$P(x) = (x - 2)(x - 3)(x + 2)$ is a cubic with positive leading coefficient, so it is strictly increasing when restricted to avoid the local maximum and minimum.

To make $f$ one-to-one, restrict to $[2,\; \infty)$ (after the local minimum at one of the turning points) or $(-\infty,\; -2]$.

Question:

The cubic curve $y = x^3 - 6x^2 + 11x - 6$ intersects the $x$-axis at points $A$, $B$ And $C$. Find the coordinates of $A$, $B$, $C$ and the area of triangle $ABC$.

Solution:

$x^3 - 6x^2 + 11x - 6 = 0$

By inspection $x = 1$: $1 - 6 + 11 - 6 = 0$. So $(x - 1)$ is a factor.

Dividing: $x^3 - 6x^2 + 11x - 6 = (x - 1)(x^2 - 5x + 6) = (x - 1)(x - 2)(x - 3)$.

Roots: $x = 1, 2, 3$.

A = (1, 0)$$B = (2, 0)$$C = (3, 0).

Since all three points lie on the $x$-axis, they are collinear, and the area of triangle $ABC$ is $0$.

Question:

Given that x - 1$$x + 2 And $x - 3$ are factors of $P(x) = x^3 + ax^2 + bx + c$Find a$$b And $c$.

Solution:

Since x - 1$$x + 2 And $x - 3$ are all factors of the cubic $P(x)$We can write:

$P(x) = (x - 1)(x + 2)(x - 3)$

Expanding: $(x - 1)(x^2 - x - 6) = x^3 - x^2 - 6x - x^2 + x + 6 = x^3 - 2x^2 - 5x + 6$.

Therefore a = -2$$b = -5$$c = 6.

Question:

When $P(x) = x^3 + 2x^2 - 5x + 1$ is divided by $x^2 - x - 2$Find the quotient and remainder.

Solution:

Since we divide a cubic by a quadratic, the remainder has degree at most 1: $R(x) = Ax + B$.

$P(x) = Q(x)(x^2 - x - 2) + Ax + B$

Factorising: $x^2 - x - 2 = (x - 2)(x + 1)$.

$P(2) = 8 + 8 - 10 + 1 = 7 = A(2) + B = 2A + B$. … (1)

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

(1) - (2): $3A = 0 \implies A = 0$. Then $B = 7$.

Remainder $= 7$.

For the quotient: $P(x) - 7 = x^3 + 2x^2 - 5x - 6$.

Dividing by $x^2 - x - 2$: the leading term is $x$Giving $x(x^2 - x - 2) = x^3 - x^2 - 2x$.

$P(x) - 7 - x(x^2 - x - 2) = 3x^2 - 3x - 6 = 3(x^2 - x - 2)$.

So $Q(x) = x + 3$Remainder $= 7$.

Question:

Fully factorise $P(x) = 2x^3 - x^2 - 13x - 6$.

Solution:

By the factor theorem, try integer factors of $-6$ divided by factors of $2$: $\pm 1, \pm 2, \pm 3, \pm 6, \pm \dfrac{1}{2}, \pm \dfrac{3}{2}$.

$P(3) = 2(27) - 9 - 39 - 6 = 54 - 54 = 0$. So $(x - 3)$ is a factor.

Divide $2x^3 - x^2 - 13x - 6$ by $(x - 3)$:

$P(x) = (x - 3)(2x^2 + 5x + 2) = (x - 3)(2x + 1)(x + 2)$.

Question:

If \alpha$$\beta$$\gamma are the roots of $2x^3 - 3x^2 + 4x - 5 = 0$Find $\dfrac{1}{\alpha} + \dfrac{1}{\beta} + \dfrac{1}{\gamma}$.

Solution:

By Vieta’s formulas (for $ax^3 + bx^2 + cx + d = 0$):

\alpha + \beta + \gamma = \dfrac{3}{2}$$\alpha\beta + \beta\gamma + \gamma\alpha = 2$$\alpha\beta\gamma = \dfrac{5}{2}.

$\frac{1}{\alpha} + \frac{1}{\beta} + \frac{1}{\gamma} = \frac{\alpha\beta + \beta\gamma + \gamma\alpha}{\alpha\beta\gamma} = \frac{2}{5/2} = \frac{4}{5}$

Question:

Find the remainder when $x^{100} + x^{50} + 1$ is divided by $x^2 - 1$.

Solution:

$x^2 - 1 = (x - 1)(x + 1)$.

Remainder has the form $Ax + B$.

At $x = 1$: $P(1) = 1 + 1 + 1 = 3 = A + B$. … (1)

At $x = -1$: $P(-1) = 1 + 1 + 1 = 3 = -A + B$. … (2)

(1) + (2): 2B = 6 \implies B = 3$$A = 0.

Remainder $= 3$.

Question:

Find constants A$$B$$C such that $\dfrac{3x + 7}{(x + 1)(x + 2)} = \dfrac{A}{x + 1} + \dfrac{B}{x + 2}$.

Solution:

$3x + 7 = A(x + 2) + B(x + 1)$

At $x = -1$: $-3 + 7 = A(1) + 0 \implies A = 4$.

At $x = -2$: $-6 + 7 = 0 + B(-1) \implies B = -1$.

Verification: $\dfrac{4}{x+1} - \dfrac{1}{x+2} = \dfrac{4(x+2) - (x+1)}{(x+1)(x+2)} = \dfrac{4x + 8 - x - 1}{(x+1)(x+2)} = \dfrac{3x + 7}{(x+1)(x+2)}$. Correct.

Question:

If $\alpha$ and $\beta$ are the roots of $2x^2 + 3x - 4 = 0$Find the equation whose roots are $\dfrac{1}{\alpha^2}$ and $\dfrac{1}{\beta^2}$.

Solution:

$\alpha + \beta = -\dfrac{3}{2}$, $\alpha\beta = -2$.

$\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta = \frac{9}{4} + 4 = \frac{25}{4}$

$\alpha^2 \beta^2 = 4$

$\frac{1}{\alpha^2} + \frac{1}{\beta^2} = \frac{\alpha^2 + \beta^2}{\alpha^2 \beta^2} = \frac{25/4}{4} = \frac{25}{16}$

$\frac{1}{\alpha^2} \cdot \frac{1}{\beta^2} = \frac{1}{4}$

Required equation: $x^2 - \dfrac{25}{16}x + \dfrac{1}{4} = 0$I.e. $16x^2 - 25x + 4 = 0$.

Question:

For $P(x) = -x^4 + 4x^2 - 3$Find:

(a) The $x$-intercepts. (b) The $y$-intercept. (c) The maximum value of $P(x)$.

Solution:

(a) $-x^4 + 4x^2 - 3 = 0 \implies x^4 - 4x^2 + 3 = 0$.

Let $u = x^2$: $u^2 - 4u + 3 = 0 \implies (u-1)(u-3) = 0$.

$u = 1 \implies x = \pm 1$. $u = 3 \implies x = \pm\sqrt{3}$.

$x$-intercepts: $(-\sqrt{3}, 0), (-1, 0), (1, 0), (\sqrt{3}, 0)$.

(b) $P(0) = -3$. $y$-intercept: $(0, -3)$.

(c) Let $v = x^2 \geq 0$. $P(x) = -(v^2 - 4v + 3) = -(v - 2)^2 + 4 - 3 = -(v-2)^2 + 1$.

Maximum is $1$ when $v = 2$I.e. $x^2 = 2$ So $x = \pm\sqrt{2}$.

flowchart TD
    A[Diag Polynomials] --> B[Key Concepts]
    A --> C[Core Principles]
    A --> D[Practical Applications]
    B --> E[Fundamental definitions]
    C --> F[Design patterns]
    D --> G[Real-world usage]

A factory assembly line: Polynomials are like expressions built from repeated operations — the factor theorem tells you which “ingredients” (factors) go into making the polynomial, and division strips away known factors to find what’s left.

Why it matters: Factoring polynomials is the backbone of solving equations, simplifying expressions, and understanding function behavior. Vieta’s formulas connect roots to coefficients without solving.

The key insight: (x - a) is a factor if and only if P(a) = 0 — this simple test replaces long division and reveals hidden structure.

1. Testing the wrong value in the factor theorem. For the factor $(x - a)$You must evaluate $P(a)$Not $P(-a)$. For $(x + a)$Evaluate $P(-a)$. The sign is the most common source of error in factor theorem problems.

2. Not verifying the factorisation. After polynomial division, always expand the quotient times divisor plus remainder to verify you get back the original polynomial. This catches arithmetic errors.

3. Incorrect Vieta’s sign conventions. For $ax^3 + bx^2 + cx + d = 0$: sum of roots $= -b/a$Sum of pairwise products $= c/a$Product $= -d/a$. The alternating signs are easy to confuse.

4. Assuming a polynomial has rational roots. Not all polynomials factorise with rational roots. If the rational root theorem yields no valid candidates, the polynomial may have irrational or complex roots.

5. Forgetting the degree of the remainder. When dividing by a polynomial of degree $m$The remainder has degree less than $m$. Dividing by a quadratic gives a linear (or constant) remainder, not a quadratic one.

Let $P(x) = x^3 - 4x^2 + x + 6$.

(a) Show that $(x + 1)$ is a factor of $P(x)$. (1 mark) (b) Hence factorise $P(x)$ completely. (3 marks) (c) Solve $P(x) = 0$. (1 mark) (d) Sketch the graph of $y = P(x)$Indicating the $x$-intercepts and the $y$-intercept. (3 marks)

Solution:

(a) $P(-1) = -1 - 4 - 1 + 6 = 0$. Confirmed.

(b) Divide by $(x + 1)$: $x^3 - 4x^2 + x + 6 = (x+1)(x^2 - 5x + 6) = (x+1)(x-2)(x-3)$.

(c) $x = -1$, $x = 2$, $x = 3$.

(d) $y$-intercept: $(0, 6)$. $x$-intercepts: (-1, 0)$$(2, 0)$$(3, 0). The cubic has positive leading coefficient, so it goes from bottom-left to top-right, crossing the $x$-axis at each root.

When $P(x) = 2x^3 + px^2 + qx + 3$ is divided by $(x - 1)$The remainder is $6$. When divided by $(x + 2)$The remainder is $-15$.

(a) Find $p$ and $q$. (4 marks) (b) Find the remainder when $P(x)$ is divided by $(x - 2)(x + 1)$. (3 marks)

Solution:

(a) $P(1) = 2 + p + q + 3 = 6 \implies p + q = 1$. … (1)

$P(-2) = -16 + 4p - 2q + 3 = -15 \implies 4p - 2q = -2 \implies 2p - q = -1$. … (2)

(1) + (2): 3p = 0 \implies p = 0$$q = 1.

(b) $P(x) = 2x^3 + x + 3$.

Remainder when divided by $(x-2)(x+1)$: let $R(x) = Ax + B$.

$P(2) = 16 + 2 + 3 = 21 = 2A + B$. … (1)

$P(-1) = -2 - 1 + 3 = 0 = -A + B$. … (2)

(1) + (2): 3B = 21 \implies B = 7$$A = 7.

Remainder $= 7x + 7$.

The equation $x^3 + ax^2 + bx + c = 0$ has roots \alpha$$\beta$$\gamma where \alpha + \beta + \gamma = 6$$\alpha\beta + \beta\gamma + \gamma\alpha = 11 And $\alpha\beta\gamma = 6$.

(a) Find a$$b And $c$. (2 marks) (b) Find the values of $\alpha^2 + \beta^2 + \gamma^2$. (2 marks) (c) Find the equation whose roots are \alpha + 1$$\beta + 1$$\gamma + 1. (3 marks)

Solution:

(a) a = -6$$b = 11$$c = -6.

Equation: $x^3 - 6x^2 + 11x - 6 = 0 = (x-1)(x-2)(x-3)$.

(b) $\alpha^2 + \beta^2 + \gamma^2 = (\alpha + \beta + \gamma)^2 - 2(\alpha\beta + \beta\gamma + \gamma\alpha) = 36 - 22 = 14$.

(c) Sum of new roots: $(\alpha+1) + (\beta+1) + (\gamma+1) = 6 + 3 = 9$.

Sum of pairwise products: $(\alpha+1)(\beta+1) + (\beta+1)(\gamma+1) + (\gamma+1)(\alpha+1) = (\alpha\beta + \alpha + \beta + 1) + \ldots = 11 + 2 \times 6 + 3 = 26$.

Product: $(\alpha+1)(\beta+1)(\gamma+1) = \alpha\beta\gamma + (\alpha\beta + \beta\gamma + \gamma\alpha) + (\alpha + \beta + \gamma) + 1 = 6 + 11 + 6 + 1 = 24$.

Equation: $x^3 - 9x^2 + 26x - 24 = 0$.

(a) Express $\dfrac{5x - 1}{(x + 2)(2x - 1)}$ in partial fractions. (4 marks) (b) Hence find $\displaystyle\int \frac{5x - 1}{(x + 2)(2x - 1)} \, dx$. (2 marks)

Solution:

(a) $\dfrac{5x - 1}{(x + 2)(2x - 1)} = \dfrac{A}{x + 2} + \dfrac{B}{2x - 1}$.

$5x - 1 = A(2x - 1) + B(x + 2)$.

At $x = -2$: $-11 = A(-5) \implies A = \dfrac{11}{5}$.

At $x = \dfrac{1}{2}$: $\dfrac{5}{2} - 1 = B\left(\dfrac{5}{2}\right) \implies B = 1$.

$\frac{5x - 1}{(x + 2)(2x - 1)} = \frac{11/5}{x + 2} + \frac{1}{2x - 1}$

(b) $\displaystyle\int \left(\frac{11/5}{x+2} + \frac{1}{2x-1}\right) dx = \frac{11}{5}\ln|x + 2| + \frac{1}{2}\ln|2x - 1| + C$.

$P(x) = x^4 + ax^3 + bx^2 + cx + d$ has roots $1, -1, 2, -3$.

(a) Find $a$, $b$, $c$, $d$. (3 marks) (b) Find the value of $P'(1)$. (3 marks)

Solution:

(a) $P(x) = (x-1)(x+1)(x-2)(x+3) = (x^2 - 1)(x^2 + x - 6) = x^4 + x^3 - 6x^2 - x^2 - x + 6 = x^4 + x^3 - 7x^2 - x + 6$.

a = 1$$b = -7$$c = -1$$d = 6.

(b) $P'(x) = 4x^3 + 3x^2 - 14x - 1$.

$P'(1) = 4 + 3 - 14 - 1 = -8$.

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