Question: A cubic polynomial $f(x)$ has roots $x = 1$, $x = -2$, $x = 3$ and leading coefficient $2$. Express $f(x)$ in descending powers of $x$.

Step 1. Factored form:

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

Step 2. Expand pair by pair:

$$\begin{aligned}(x - 1)(x + 2) &= x^2 + x - 2 \\ (x^2 + x - 2)(x - 3) &= x^3 - 3x^2 + x^2 - 3x - 2x + 6 \\ &= x^3 - 2x^2 - 5x + 6\end{aligned}$$

Step 3. Multiply by leading coefficient:

$$f(x) = 2x^3 - 4x^2 - 10x + 12$$

Check: $f(1) = 2 - 4 - 10 + 12 = 0$ ✓.

Question: A cubic polynomial $h(x)$ has $x = 2$ as a root, has $x^2 + 1$ as a factor, and $h(0) = 4$. Find $h(x)$ in descending powers.

Step 1. With root $x = 2$ and quadratic factor $x^2 + 1$:

$$h(x) = k(x - 2)(x^2 + 1)$$

for some constant $k$.

Step 2. Use $h(0) = 4$ to find $k$:

$$k(-2)(1) = 4 \;\Rightarrow\; k = -2$$

Step 3. Expand:

$$\begin{aligned}h(x) &= -2(x - 2)(x^2 + 1) \\ &= -2(x^3 + x - 2x^2 - 2) \\ &= -2x^3 + 4x^2 - 2x + 4\end{aligned}$$

Watch out: When the leading coefficient isn't given directly, use any other known value (e.g. $h(0)$) as the extra condition to pin it down.

Question: The cubic $f(x)$ has coefficient of $x^3$ equal to $1$ and roots $-1$, $m$, $2m$, where $m$ is an integer. Given $f(x)$ has remainder $6$ when divided by $x - 1$, find $f(x)$.

Solution: Roots $\Rightarrow$ factors are $(x + 1)$, $(x - m)$, $(x - 2m)$. Since the leading coefficient is $1$:

$$f(x) = (x + 1)(x - m)(x - 2m)$$

Use $f(1) = 6$:

$$(2)(1 - m)(1 - 2m) = 6 \Rightarrow (1 - m)(1 - 2m) = 3$$

Expand: $1 - 2m - m + 2m^2 = 3 \Rightarrow 2m^2 - 3m - 2 = 0$.

Factorise: $(m - 2)(2m + 1) = 0 \Rightarrow m = 2 \text{ or } m = -\tfrac{1}{2}$.

Since $m$ is an integer, $m = 2$.

$$\boxed{f(x) = (x + 1)(x - 2)(x - 4)}$$

Question: $f(x)$ has highest-power term $6x^3$. One root of $f(x) = 0$ is $k$, and $2x^2 + 3x + 4$ is a quadratic factor of $f(x)$. $f(x)$ leaves remainder $108$ when divided by $x - 1$.

(i) Show $k = -3$. (ii) Find the number of real roots of $f(x) = 0$.

Solution:

(i) Since $k$ is a root, $(x - k)$ is a factor. Combine with the quadratic factor:

$$f(x) = 3(x - k)(2x^2 + 3x + 4)$$

(coefficient $3$ from $3 \cdot 2 = 6$). Use $f(1) = 108$:

$$3(1 - k)(2 + 3 + 4) = 108 \Rightarrow 3(1 - k)(9) = 108 \Rightarrow 1 - k = 4 \Rightarrow \boxed{k = -3}$$

(ii) $f(x) = 0 \Rightarrow 3(x + 3)(2x^2 + 3x + 4) = 0$.

Linear factor: $x = -3$. Quadratic factor has discriminant $3^2 - 4(2)(4) = 9 - 32 = -23 < 0$ — no real roots.

$\therefore$ 1 real root.

Question: Find the remainder when $P(x) = 8x^3 + 22x^2 + 11x - 7$ is divided by: (a) $x - 1$ (b) $2x + 3$

(a) Remainder theorem: divisor $x - 1 \Rightarrow$ substitute $x = 1$.

$$P(1) = 8 + 22 + 11 - 7 = 34$$

Remainder $= 34$.

(b) Divisor $2x + 3 \Rightarrow$ substitute $x = -\dfrac{3}{2}$.

$$\begin{aligned}P\!\left(-\tfrac{3}{2}\right) &= 8\!\left(-\tfrac{3}{2}\right)^{\!3} + 22\!\left(-\tfrac{3}{2}\right)^{\!2} + 11\!\left(-\tfrac{3}{2}\right) - 7 \\ &= -27 + \tfrac{99}{2} - \tfrac{33}{2} - 7 \\ &= -1\end{aligned}$$

Remainder $= -1$.

⚠ Watch out: For divisor $ax + b$, substitute $x = -\dfrac{b}{a}$ — not just $x = -b$.

Question: $P(x) = 2x^3 + ax^2 + bx - 30$ has $(x+3)$ as a factor and leaves remainder $-28$ when divided by $(x-1)$. Find $a$ and $b$.

Step 1. Factor theorem: $P(-3) = 0$.

$$2(-27) + 9a - 3b - 30 = 0 \;\Rightarrow\; 3a - b = 28 \quad (1)$$

Step 2. Remainder theorem: $P(1) = -28$.

$$2 + a + b - 30 = -28 \;\Rightarrow\; a + b = 0 \quad (2)$$

Step 3. From (2), $b = -a$. Sub into (1):

$$3a + a = 28 \;\Rightarrow\; a = 7, \; b = -7$$

Tip: Factor gives $P(\text{root}) = 0$; remainder gives $P(\text{root}) = R$. Two conditions, two equations — always solvable when you have two unknowns.

Question: A polynomial $f(x)$ has highest term $6x^3$, one root $\alpha$, and $2x^2 + 3x + 4$ as a quadratic factor. The remainder when $f(x)$ is divided by $x - 1$ is $108$. Show that $\alpha = -3$.

Step 1. Leading term of $f(x)$ is $6x^3$ and the quadratic factor has leading $2x^2$, so the linear factor must have leading $3x$:

$$f(x) = 3(x - \alpha)(2x^2 + 3x + 4)$$

Step 2. Apply $f(1) = 108$:

$$\begin{aligned}3(1 - \alpha)(2 + 3 + 4) &= 108 \\ 27(1 - \alpha) &= 108 \\ 1 - \alpha &= 4 \\ \alpha &= -3\end{aligned}$$

Why this works: Keeping $f(x)$ in factored form turns "substitute $x = 1$" into a one-line evaluation. Don't expand unless you have to.

Question: Given that $x + 4$ is a factor of $g(x) + 3$, where $g(x)$ is a polynomial, find the remainder when $f(x) = (x^3 - 3x + 2)\,g(x)$ is divided by $x + 4$.

Solution:

The factor $x + 4$ being a factor of $g(x) + 3$ means $g(-4) + 3 = 0$, so $g(-4) = -3$.

By the Remainder Theorem, the remainder when $f(x)$ is divided by $x + 4$ is $f(-4)$:

$$\begin{aligned} f(-4) &= \big((-4)^3 - 3(-4) + 2\big)\,g(-4) \\ &= (-64 + 12 + 2)(-3) \\ &= -50 \times (-3) \\ &= \boxed{150} \end{aligned}$$

⚠ Watch out: the polynomial inside the multiplied bracket is $g(x)$, not $g(x) + 3$. Substitute $g(-4) = -3$ on its own.

Question: If $x^3 - hx^2 + kx - 9$ has a factor of $x^2 + 3$, find $h$ and $k$ and the other factor.

Solution: Since $x^2 + 3$ is a factor and the polynomial is cubic, the quotient must be linear: $(ax + b)$. Write the identity (no remainder, because we have a factor):

$$x^3 - hx^2 + kx - 9 = (x^2 + 3)(ax + b)$$

Expand the RHS: $(x^2 + 3)(ax + b) = ax^3 + bx^2 + 3ax + 3b$.

Compare coefficients:

$$\begin{aligned} x^3 &: a = 1 \\ \text{const} &: 3b = -9 \Rightarrow b = -3 \\ x^2 &: -h = b = -3 \Rightarrow h = 3 \\ x &: k = 3a = 3 \end{aligned}$$

Answer: $h = 3$, $k = 3$, the other factor is $x - 3$.

Question: Factorise $f(x) = 2x^3 - 5x^2 - 4x + 3$ completely.

Step 1. Test integer-root candidates ($\pm 1, \pm 3$): $f(-1) = -2 - 5 + 4 + 3 = 0$ ✓

So $(x + 1)$ is a factor.

Step 2. Divide $f(x)$ by $(x + 1)$:

$$f(x) = (x + 1)(2x^2 - 7x + 3)$$

Step 3. Factorise the quadratic:

$$2x^2 - 7x + 3 = (2x - 1)(x - 3)$$

Final:

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

Tip: Rational-root test — try $\pm \dfrac{\text{factors of constant}}{\text{factors of leading coeff}}$. Cuts candidates down to a small finite list.

Question: Factorise $g(x) = x^4 - 5x^3 + 5x^2 + 5x - 6$ completely.

Step 1. Test $g(1) = 1 - 5 + 5 + 5 - 6 = 0$ ✓. So $(x - 1)$ is a factor.

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

Step 2. Now factorise the cubic. Test $(x + 1)$: $-1 - 4 - 1 + 6 = 0$ ✓.

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

Step 3. Factorise the quadratic:

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

Final:

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

⚠ Watch out: A quartic usually needs two rounds of factor theorem before you reach a quadratic. Don't stop at one factor.

Question: $f(x) = 3(x + 3)(2x^2 + 3x + 4)$. How many real roots does $f(x) = 0$ have?

Step 1. Set each factor to zero:

• $x + 3 = 0 \;\Rightarrow\; x = -3$
• $2x^2 + 3x + 4 = 0 \;\Rightarrow\;$ check discriminant.

Step 2. Discriminant of $2x^2 + 3x + 4$:

$$\Delta = 3^2 - 4(2)(4) = 9 - 32 = -23 < 0$$

Since $\Delta < 0$, this quadratic has no real roots.

Step 3. $f(x) = 0$ has exactly one real root: $x = -3$.

Why this works: Once a polynomial is fully factorised, every real root comes from a factor that can equal zero over the reals. An irreducible quadratic contributes zero real roots.

Question: Factorise $54x^6 - 16y^3$ completely.

Step 1. Take out the common factor:

$$54x^6 - 16y^3 = 2(27x^6 - 8y^3)$$

Step 2. Recognise the difference of cubes: $27x^6 = (3x^2)^3$ and $8y^3 = (2y)^3$.

Apply $a^3 - b^3 = (a - b)(a^2 + ab + b^2)$ with $a = 3x^2$, $b = 2y$:

$$\begin{aligned}27x^6 - 8y^3 &= (3x^2 - 2y)\!\bigl[(3x^2)^2 + (3x^2)(2y) + (2y)^2\bigr] \\ &= (3x^2 - 2y)(9x^4 + 6x^2 y + 4y^2)\end{aligned}$$

Final:

$$54x^6 - 16y^3 = 2(3x^2 - 2y)(9x^4 + 6x^2 y + 4y^2)$$

Identities to memorise:

$$a^3 - b^3 = (a - b)(a^2 + ab + b^2)$$ $$a^3 + b^3 = (a + b)(a^2 - ab + b^2)$$

Question: Factorise $(2 + x)^3 - 216$ completely.

Step 1. Rewrite $216 = 6^3$. Difference of cubes with $a = 2 + x$, $b = 6$.

Step 2. Apply $a^3 - b^3 = (a - b)(a^2 + ab + b^2)$:

$$(2 + x)^3 - 6^3 = \bigl[(2 + x) - 6\bigr]\bigl[(2 + x)^2 + 6(2 + x) + 36\bigr]$$

Step 3. Simplify each bracket:

$$\begin{aligned}(2 + x) - 6 &= x - 4 \\ (2 + x)^2 + 6(2 + x) + 36 &= 4 + 4x + x^2 + 12 + 6x + 36 \\ &= x^2 + 10x + 52\end{aligned}$$

Final:

$$(2 + x)^3 - 216 = (x - 4)(x^2 + 10x + 52)$$

Tip: Sum/difference of cubes works even when $a$ or $b$ is itself a binomial — just keep track of the brackets carefully.

Question:
(i) Factorise $(2 + x)^3 - 216$ completely.
(ii) Hence show $(2 + x)^3 - 216 = (x - 4)^2$ has exactly one real root.

Solution:

(i) Write $216 = 6^3$, then apply $a^3 - b^3 = (a - b)(a^2 + ab + b^2)$ with $a = 2 + x$, $b = 6$:

$$\begin{aligned} (2 + x)^3 - 216 &= \big[(2 + x) - 6\big]\big[(2 + x)^2 + 6(2 + x) + 36\big] \\ &= (x - 4)(x^2 + 4x + 4 + 12 + 6x + 36) \\ &= \boxed{(x - 4)(x^2 + 10x + 52)} \end{aligned}$$

(ii) Set the original equation:

$$(x - 4)(x^2 + 10x + 52) = (x - 4)^2$$ $$(x - 4)\big[(x^2 + 10x + 52) - (x - 4)\big] = 0$$ $$(x - 4)(x^2 + 9x + 56) = 0$$

So $x = 4$, or $x^2 + 9x + 56 = 0$. Discriminant $= 81 - 4(56) = -143 < 0$, no real roots.

$\therefore$ exactly one real root, $x = 4$.

Question: Find constants $a$, $b$, $c$ so that

$$2x^3 - 3x^2 + x - 5 \equiv (x - 1)(2x^2 + ax + b) + c.$$

Step 1. Expand the right-hand side:

$$\begin{aligned}\text{RHS} &= 2x^3 + ax^2 + bx - 2x^2 - ax - b + c \\ &= 2x^3 + (a - 2)x^2 + (b - a)x + (c - b)\end{aligned}$$

Step 2. Compare with LHS $2x^3 - 3x^2 + x - 5$:

• $x^2$: $\;a - 2 = -3 \Rightarrow a = -1$
• $x$: $\;b - a = 1 \Rightarrow b = 0$
• const: $\;c - b = -5 \Rightarrow c = -5$

Answer: $a = -1$, $b = 0$, $c = -5$.

Tip: $\equiv$ means equal for all $x$, so coefficients of like powers must match on both sides.

Question: $A(x - 1) + B(x + 2) \equiv 3x + 1$ for all $x$. Find $A$ and $B$.

Step 1. Substitute $x = 1$ (kills the $A$ term):

$$0 + B(3) = 3(1) + 1 = 4 \;\Rightarrow\; B = \dfrac{4}{3}$$

Step 2. Substitute $x = -2$ (kills the $B$ term):

$$A(-3) + 0 = 3(-2) + 1 = -5 \;\Rightarrow\; A = \dfrac{5}{3}$$

Answer: $A = \dfrac{5}{3}$, $B = \dfrac{4}{3}$.

Why this works: When an identity has each unknown isolated in a separate factor, pick $x$ to zero out all but one term — faster than expand-and-compare.

Question: Find the quotient and remainder when $2x^3 + 5x^2 - 2x + 7$ is divided by $x + 3$.

Solution: Long division — divide leading terms each step.

$$\begin{array}{r}2x^2 - x + 1 \phantom{)}\\ x + 3 \enclose{longdiv}{2x^3 + 5x^2 - 2x + 7}\\ \underline{2x^3 + 6x^2 \phantom{- 2x + 7}}\\ -x^2 - 2x \phantom{+ 7}\\ \underline{-x^2 - 3x \phantom{+ 7}}\\ x + 7 \\ \underline{x + 3}\\ 4\end{array}$$

Quotient: $2x^2 - x + 1$. Remainder: $4$.

Check: $(x + 3)(2x^2 - x + 1) + 4 = 2x^3 + 5x^2 - 2x + 7$ ✓.

Tip: Always multiply back to verify — saves marks lost to sign errors.

Question: Find the quotient and remainder when $3x^4 - 2x^2 + 5$ is divided by $x^2 - 2$.

Solution: Insert placeholders for missing degrees so columns align:

$$3x^4 + 0x^3 - 2x^2 + 0x + 5$$

Long-divide by $x^2 - 2$:

$$\begin{aligned}\text{Step 1:}\; & 3x^4 \div x^2 = 3x^2. \\ &3x^2 \cdot (x^2 - 2) = 3x^4 - 6x^2. \\ &\text{Subtract: } -2x^2 - (-6x^2) = 4x^2. \\ \text{Step 2:}\; & 4x^2 \div x^2 = 4. \\ &4 \cdot (x^2 - 2) = 4x^2 - 8. \\ &\text{Subtract: } 5 - (-8) = 13.\end{aligned}$$

Quotient: $3x^2 + 4$. Remainder: $13$.

⚠ Watch out: Include $0x^3$ and $0x$ placeholders for missing-degree terms — easy to drop a term and end up with a wrong remainder otherwise.

Question: When the polynomial $x^{10} - px^3 + q$ is divided by $x^2 - 1$, the remainder is $4x + 3$. Find $p$ and $q$. State the expression to be added so that $x^{10} - px^3 + q$ is completely divisible by $x^2 - 1$.

Solution: Use $f(x) = d(x)\,Q(x) + r(x)$ with $r(x) = 4x + 3$ (remainder is linear since divisor is quadratic). Factor $x^2 - 1 = (x - 1)(x + 1)$. Substitute the roots of the divisor:

$$\begin{aligned} x = 1 &: 1 - p + q = 4 + 3 \Rightarrow -p + q = 6 \;\;(1) \\ x = -1 &: 1 + p + q = -4 + 3 \Rightarrow p + q = -2 \;\;(2) \end{aligned}$$

Add: $2q = 4 \Rightarrow q = 2$. Then $p = -4$.

To make it exactly divisible, kill the remainder $4x + 3$ by adding $-(4x + 3)$.

$\therefore$ Add $-(4x + 3)$ — i.e. subtract $4x + 3$.

Question: Solve $2x^3 + x^2 - 7x + 4 = 0$, giving non-integer roots in surd form.

Step 1. Try $f(1) = 2 + 1 - 7 + 4 = 0$ ✓. So $(x - 1)$ is a factor.

Divide: $2x^3 + x^2 - 7x + 4 = (x - 1)(2x^2 + 3x - 4)$.

Step 2. Solve each factor.

• $x - 1 = 0 \Rightarrow x = 1$.
• $2x^2 + 3x - 4 = 0$ — use the quadratic formula:

$$x = \dfrac{-3 \pm \sqrt{9 + 32}}{4} = \dfrac{-3 \pm \sqrt{41}}{4}$$

Step 3. All roots:

$$x = 1, \quad x = \dfrac{-3 + \sqrt{41}}{4}, \quad x = \dfrac{-3 - \sqrt{41}}{4}$$

⚠ Watch out: "Surd form" means keep $\sqrt{41}$ exact — don't decimal-approximate.

Question: Given $P(x) = 2x^3 + 7x^2 - 7x - 30$ has $(x + 3)$ as a factor. Solve $P(x) = 0$.

Step 1. Divide $P(x)$ by $(x + 3)$:

$$P(x) = (x + 3)(2x^2 + x - 10)$$

Step 2. Factorise the quadratic:

$$2x^2 + x - 10 = (2x + 5)(x - 2)$$

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

Step 3. Set each factor to zero:

$$x = -3, \quad x = -\dfrac{5}{2}, \quad x = 2$$

Tip: When "Hence solve" follows a "Find $a$, $b$" part, you've usually already done the hard work — the factor used in the factor-theorem step is one of the roots.