Polynomials — Diagnostic Tests
Unit Tests
Tests edge cases, boundary conditions, and common misconceptions for polynomials.
UT-1: Factor Theorem Sign Confusion
Question:
Determine whether is a factor of .
Solution:
By the factor theorem, is a factor if and only if .
Since , is not a factor.
A common mistake is testing instead of . The factor theorem states that is a factor if . Here the factor is , so we test .
UT-2: Remainder Theorem Application
Question:
When is divided by , the remainder is . When is divided by , the remainder is . Find and (if also has a constant term correction replacing ), or simply find .
Solution:
By the remainder theorem:
.
Verify: ?
Wait: .
But we need . There is no single value of satisfying both conditions simultaneously with the constant term fixed at . The problem as stated is inconsistent. This highlights the importance of verifying all conditions.
If instead the constant term is also a variable :
.
.
Subtracting: , .
UT-3: Polynomial Division with Errors
Question:
Divide by .
Solution:
x^2 - 2x + 3 & 2x^3 - 5x^2 + \phantom{0}x - 6 \\ \hline & 2x \\ & 2x^3 - 4x^2 + 6x \\ \hline & \phantom{2x^3} - x^2 - 5x - 6 \\ & \phantom{2x^3} - x^2 + 2x - 3 \\ \hline & \phantom{2x^3} \phantom{- x^2} - 7x - 3 \\ \end{array}$$ Quotient: $2x - 1$, Remainder: $-7x - 3$. Verify: $(2x - 1)(x^2 - 2x + 3) + (-7x - 3)$ $= 2x^3 - 4x^2 + 6x - x^2 + 2x - 3 - 7x - 3$ $= 2x^3 - 5x^2 + x - 6$. Correct. --- ### UT-4: Vieta's Formulas for Cubic **Question:** If $\alpha$, $\beta$, $\gamma$ are the roots of $x^3 - 5x^2 + 2x + 8 = 0$, find: (a) $\alpha + \beta + \gamma$ (b) $\alpha\beta + \beta\gamma + \gamma\alpha$ (c) $\alpha\beta\gamma$ **Solution:** By Vieta's formulas for $x^3 + px^2 + qx + r = 0$: $$\alpha + \beta + \gamma = -p = 5$$ $$\alpha\beta + \beta\gamma + \gamma\alpha = q = 2$$ $$\alpha\beta\gamma = -r = -8$$ --- ### UT-5: Finding Unknown Coefficients Given Factors **Question:** $P(x) = x^3 + ax^2 + bx - 12$ has factors $(x - 1)$ and $(x + 4)$. Find $a$, $b$, and the remaining factor. **Solution:** Since $(x - 1)$ is a factor: $P(1) = 1 + a + b - 12 = 0 \implies a + b = 11$. ... (1) Since $(x + 4)$ is a factor: $P(-4) = -64 + 16a - 4b - 12 = 0 \implies 16a - 4b = 76 \implies 4a - b = 19$. ... (2) From (1) + (2): $5a = 30 \implies a = 6$. Then $b = 5$. So $P(x) = x^3 + 6x^2 + 5x - 12$. Dividing by $(x - 1)(x + 4) = x^2 + 3x - 4$: $x^3 + 6x^2 + 5x - 12 = (x^2 + 3x - 4)(x + c)$ Expanding RHS: $x^3 + cx^2 + 3x^2 + 3cx - 4x - 4c = x^3 + (c+3)x^2 + (3c-4)x - 4c$. Matching: $c + 3 = 6 \implies c = 3$. The remaining factor is $(x + 3)$. --- ## Integration Tests > Tests synthesis of polynomials with other topics. ### IT-1: Polynomials and Inequalities (with Inequalities) **Question:** Let $P(x) = (x - 1)(x^2 - 4x + 3)$. Find the set of values of $x$ for which $P(x) \leq 0$. **Solution:** $$P(x) = (x - 1)(x - 1)(x - 3) = (x - 1)^2(x - 3)$$ Critical values: $x = 1$ (double root) and $x = 3$. | Interval | Test | $(x-1)^2$ | $(x-3)$ | Product | |---|---|---|---|---| | $x < 1$ | $x = 0$ | $+$ | $-$ | $-$ | | $1 < x < 3$ | $x = 2$ | $+$ | $-$ | $-$ | | $x > 3$ | $x = 4$ | $+$ | $+$ | $+$ | $P(x) \leq 0$ when $x \leq 3$ (including $x = 1$ and $x = 3$). Therefore $x \in (-\infty,\; 3]$. --- ### IT-2: Polynomials and Functions (with Functions) **Question:** Let $f(x) = x^3 - 3x^2 - 4x + 12$. Given that $(x - 2)$ is a factor, find all $x$ for which $f(x) = 0$, and hence state the domain on which $f$ is one-to-one. **Solution:** $P(2) = 8 - 12 - 8 + 12 = 0$. Confirmed. Divide $x^3 - 3x^2 - 4x + 12$ by $(x - 2)$: $$\begin{array}{r|l} x - 2 & x^3 - 3x^2 - 4x + 12 \\ \hline & x^2 - x - 6 \\ & x^3 - 2x^2 \\ \hline & \phantom{x^3} - x^2 - 4x \\ & \phantom{x^3} - x^2 + 2x \\ \hline & \phantom{x^3 x^2} - 6x + 12 \\ & \phantom{x^3 x^2} - 6x + 12 \\ \hline & \phantom{x^3 x^2} 0 \\ \end{array}$$ $$x^2 - x - 6 = (x - 3)(x + 2)$$ Roots: $x = -2$, $x = 2$, $x = 3$. $P(x) = (x - 2)(x - 3)(x + 2)$ is a cubic with positive leading coefficient, so it is strictly increasing when restricted to avoid the local maximum and minimum. To make $f$ one-to-one, restrict to $[2,\; \infty)$ (after the local minimum at one of the turning points) or $(-\infty,\; -2]$. --- ### IT-3: Polynomials and Coordinate Geometry (with Coordinate Geometry) **Question:** The cubic curve $y = x^3 - 6x^2 + 11x - 6$ intersects the $x$-axis at points $A$, $B$, and $C$. Find the coordinates of $A$, $B$, $C$ and the area of triangle $ABC$. **Solution:** $x^3 - 6x^2 + 11x - 6 = 0$ By inspection $x = 1$: $1 - 6 + 11 - 6 = 0$. So $(x - 1)$ is a factor. Dividing: $x^3 - 6x^2 + 11x - 6 = (x - 1)(x^2 - 5x + 6) = (x - 1)(x - 2)(x - 3)$. Roots: $x = 1, 2, 3$. $A = (1, 0)$, $B = (2, 0)$, $C = (3, 0)$. Since all three points lie on the $x$-axis, they are collinear, and the area of triangle $ABC$ is $0$. --- ## Worked Examples ### WE-1: Factor Theorem with Multiple Factors **Question:** Given that $x - 1$, $x + 2$, and $x - 3$ are factors of $P(x) = x^3 + ax^2 + bx + c$, find $a$, $b$, and $c$. **Solution:** Since $x - 1$, $x + 2$, and $x - 3$ are all factors of the cubic $P(x)$, we can write: $$P(x) = (x - 1)(x + 2)(x - 3)$$ Expanding: $(x - 1)(x^2 - x - 6) = x^3 - x^2 - 6x - x^2 + x + 6 = x^3 - 2x^2 - 5x + 6$. Therefore $a = -2$, $b = -5$, $c = 6$. --- ### WE-2: Remainder When Dividing by Quadratic **Question:** When $P(x) = x^3 + 2x^2 - 5x + 1$ is divided by $x^2 - x - 2$, find the quotient and remainder. **Solution:** Since we divide a cubic by a quadratic, the remainder has degree at most 1: $R(x) = Ax + B$. $$P(x) = Q(x)(x^2 - x - 2) + Ax + B$$ Factorising: $x^2 - x - 2 = (x - 2)(x + 1)$. $P(2) = 8 + 8 - 10 + 1 = 7 = A(2) + B = 2A + B$. ... (1) $P(-1) = -1 + 2 + 5 + 1 = 7 = A(-1) + B = -A + B$. ... (2) (1) - (2): $3A = 0 \implies A = 0$. Then $B = 7$. Remainder $= 7$. For the quotient: $P(x) - 7 = x^3 + 2x^2 - 5x - 6$. Dividing by $x^2 - x - 2$: the leading term is $x$, giving $x(x^2 - x - 2) = x^3 - x^2 - 2x$. $P(x) - 7 - x(x^2 - x - 2) = 3x^2 - 3x - 6 = 3(x^2 - x - 2)$. So $Q(x) = x + 3$, Remainder $= 7$. --- ### WE-3: Using Factor Theorem to Fully Factorise **Question:** Fully factorise $P(x) = 2x^3 - x^2 - 13x - 6$. **Solution:** By the factor theorem, try integer factors of $-6$ divided by factors of $2$: $\pm 1, \pm 2, \pm 3, \pm 6, \pm \dfrac{1}{2}, \pm \dfrac{3}{2}$. $P(3) = 2(27) - 9 - 39 - 6 = 54 - 54 = 0$. So $(x - 3)$ is a factor. Divide $2x^3 - x^2 - 13x - 6$ by $(x - 3)$: $$\begin{array}{r|l} x - 3 & 2x^3 - x^2 - 13x - 6 \\ \hline & 2x^2 \\ & 2x^3 - 6x^2 \\ \hline & \phantom{2x^3} 5x^2 - 13x \\ & \phantom{2x^3} 5x^2 - 15x \\ \hline & \phantom{2x^3 x^2} 2x - 6 \\ & \phantom{2x^3 x^2} 2x - 6 \\ \hline & \phantom{2x^3 x^2} 0 \\ \end{array}$$ $P(x) = (x - 3)(2x^2 + 5x + 2) = (x - 3)(2x + 1)(x + 2)$. --- ### WE-4: Symmetric Sums of Roots **Question:** If $\alpha$, $\beta$, $\gamma$ are the roots of $2x^3 - 3x^2 + 4x - 5 = 0$, find $\dfrac{1}{\alpha} + \dfrac{1}{\beta} + \dfrac{1}{\gamma}$. **Solution:** By Vieta's formulas (for $ax^3 + bx^2 + cx + d = 0$): $\alpha + \beta + \gamma = \dfrac{3}{2}$, $\alpha\beta + \beta\gamma + \gamma\alpha = 2$, $\alpha\beta\gamma = \dfrac{5}{2}$. $$\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}$$ --- ### WE-5: Finding the Remainder Without Division **Question:** Find the remainder when $x^{100} + x^{50} + 1$ is divided by $x^2 - 1$. **Solution:** $x^2 - 1 = (x - 1)(x + 1)$. Remainder has the form $Ax + B$. At $x = 1$: $P(1) = 1 + 1 + 1 = 3 = A + B$. ... (1) At $x = -1$: $P(-1) = 1 + 1 + 1 = 3 = -A + B$. ... (2) (1) + (2): $2B = 6 \implies B = 3$, $A = 0$. Remainder $= 3$. --- ### WE-6: Polynomial Identity Method **Question:** Find constants $A$, $B$, $C$ such that $\dfrac{3x + 7}{(x + 1)(x + 2)} = \dfrac{A}{x + 1} + \dfrac{B}{x + 2}$. **Solution:** $$3x + 7 = A(x + 2) + B(x + 1)$$ At $x = -1$: $-3 + 7 = A(1) + 0 \implies A = 4$. At $x = -2$: $-6 + 7 = 0 + B(-1) \implies B = -1$. Verification: $\dfrac{4}{x+1} - \dfrac{1}{x+2} = \dfrac{4(x+2) - (x+1)}{(x+1)(x+2)} = \dfrac{4x + 8 - x - 1}{(x+1)(x+2)} = \dfrac{3x + 7}{(x+1)(x+2)}$. Correct. --- ### WE-7: Reciprocal Equation Roots **Question:** If $\alpha$ and $\beta$ are the roots of $2x^2 + 3x - 4 = 0$, find the equation whose roots are $\dfrac{1}{\alpha^2}$ and $\dfrac{1}{\beta^2}$. **Solution:** $\alpha + \beta = -\dfrac{3}{2}$, $\alpha\beta = -2$. $$\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$. --- ### WE-8: Polynomial Graph Behaviour **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}$. --- ## Common Pitfalls 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. --- ## DSE Exam-Style Questions ### DSE-1 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. --- ### DSE-2 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$. --- ### DSE-3 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$. --- ### DSE-4 (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) $\displaystyle\int \left(\frac{11/5}{x+2} + \frac{1}{2x-1}\right) dx = \frac{11}{5}\ln|x + 2| + \frac{1}{2}\ln|2x - 1| + C$. --- ### DSE-5 $P(x) = x^4 + ax^3 + bx^2 + cx + d$ 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) $P'(x) = 4x^3 + 3x^2 - 14x - 1$. $P'(1) = 4 + 3 - 14 - 1 = -8$.