2.P — Worked Problems: Algebra

Fifteen problems, each solved line by line with the identity being used named at every step. Each identity is derived in the chapter that introduces it.

Problem 1 — Factorising with the cube identities

Factorise completely: (a) 8x^3 + 27, (b) x^6 - 64.

Solution

(a) 8x^3 + 27.

Step 1 — get both terms into the form (something)³. 8x^3 = (2x)^3 because 2^3 = 8 and (x)^3 = x^3. And 27 = 3^3. So

8x^3 + 27 = (2x)^3 + 3^3

Step 2 — apply a^3+b^3 = (a+b)(a^2-ab+b^2) with a = 2x and b = 3:

= (2x + 3)\left((2x)^2 - (2x)(3) + 3^2\right)

Step 3 — tidy the second bracket.

= (2x+3)(4x^2 - 6x + 9)

Step 4 — can the quadratic factor further? Check its discriminant: \Delta = (-6)^2 - 4(4)(9) = 36 - 144 = -108, which is negative. No real roots, so it does not factorise over the real numbers.

Answer: (2x+3)(4x^2-6x+9).

(b) x^6 - 64. This one can be attacked two ways, and it is worth seeing both.

Route 1 — as a difference of squares first. x^6 = (x^3)^2 and 64 = 8^2:

x^6 - 64 = (x^3-8)(x^3+8)

Now each bracket is a cube difference and a cube sum:

x^3 - 8 = (x-2)(x^2+2x+4), \qquad x^3+8 = (x+2)(x^2-2x+4)

x^6 - 64 = (x-2)(x+2)(x^2+2x+4)(x^2-2x+4)

Route 2 — as a difference of cubes first. x^6 = (x^2)^3 and 64 = 4^3:

x^6 - 64 = (x^2-4)(x^4+4x^2+16)

The first bracket factors as (x-2)(x+2). The second needs a trick: add and subtract 4x^2 to make a square appear.

x^4 + 4x^2 + 16 = (x^4 + 8x^2 + 16) - 4x^2 = (x^2+4)^2 - (2x)^2 = (x^2+4-2x)(x^2+4+2x)

which lands on the same four factors.

Answer: (x-2)(x+2)(x^2+2x+4)(x^2-2x+4).

What to notice. Squares first is the easier route, and the general lesson is that when an expression is both a perfect square and a perfect cube, splitting the square first leaves smaller pieces to handle.

Problem 2 — Completing the square, and reading the answer off

Find the minimum value of f(x) = 3x^2 - 12x + 7, and say where it happens — without calculus.

Solution

Step 1 — take the coefficient of x^2 outside the first two terms. Do not touch the constant yet.

f(x) = 3\left(x^2 - 4x\right) + 7

Check: 3 \times x^2 = 3x^2 ✓, 3 \times (-4x) = -12x ✓.

Step 2 — complete the square inside the bracket. For x^2 - 4x, halve the coefficient of x to get -2, then square it to get 4. So x^2-4x+4 = (x-2)^2. We need that +4, but we cannot simply add it, so we add and subtract it in the same breath:

x^2 - 4x = (x-2)^2 - 4

Step 3 — put that back, and be careful with the 3 outside.

f(x) = 3\left[(x-2)^2 - 4\right] + 7 = 3(x-2)^2 - 12 + 7 = 3(x-2)^2 - 5

The -4 inside became -12 outside, because the 3 multiplies everything in the bracket. Forgetting this is the standard slip.

Step 4 — read the answer. A square is never negative, so (x-2)^2 \ge 0 always, and therefore 3(x-2)^2 \ge 0. The smallest f can be is when that term is exactly zero, which happens when x = 2.

Answer: the minimum value is -5, at x = 2.

Verify: f(2) = 3(4) - 24 + 7 = 12 - 24 + 7 = -5 ✓. And f(1) = 3 - 12 + 7 = -2, which is bigger ✓.

Problem 3 — A quadratic where the discriminant is the whole question

For which values of k does x^2 + kx + 9 = 0 have (a) two distinct real roots, (b) exactly one root, (c) no real roots?

Solution

Step 1 — write the discriminant. With a=1, b=k, c=9:

\Delta = b^2 - 4ac = k^2 - 36

Step 2 — translate each case into a condition on k.

(a) Two distinct real roots needs \Delta > 0:

k^2 - 36 > 0 \quad \Rightarrow \quad k^2 > 36

Careful here. k^2 > 36 does not mean k > 6. It means k is further than 6 from zero, in either direction, since squaring destroys the sign. Using the absolute-value rule |k| > 6:

k < -6 \quad\text{or}\quad k > 6

(b) Exactly one root needs \Delta = 0:

k^2 = 36 \quad \Rightarrow \quad k = 6 \text{ or } k = -6

(c) No real roots needs \Delta < 0:

k^2 < 36 \quad \Rightarrow \quad -6 < k < 6

Answers: (a) k<-6 or k>6; (b) k = \pm6; (c) -6<k<6.

Check one. Take k = 0, inside case (c). The equation is x^2 + 9 = 0, so x^2 = -9 — impossible for a real number ✓.

What to notice. Case (b) is the boundary, and boundaries are where the interesting mathematics lives: k = \pm 6 gives x^2 \pm 6x + 9 = (x\pm3)^2, a perfect square whose graph just touches the axis.

Problem 4 — Vieta without solving

The roots of 2x^2 - 7x + 3 = 0 are \alpha and \beta. Find \alpha^2 + \beta^2 and \dfrac1\alpha + \dfrac1\beta without finding the roots themselves.

Solution

Step 1 — read off the sum and product using \alpha+\beta = -b/a and \alpha\beta = c/a:

\alpha + \beta = \frac{7}{2}, \qquad \alpha\beta = \frac{3}{2}

Step 2 — express \alpha^2+\beta^2 in terms of those two. The identity (\alpha+\beta)^2 = \alpha^2 + 2\alpha\beta + \beta^2 rearranges to

\alpha^2 + \beta^2 = (\alpha+\beta)^2 - 2\alpha\beta

Step 3 — substitute.

= \left(\frac72\right)^2 - 2\left(\frac32\right) = \frac{49}{4} - 3 = \frac{49 - 12}{4} = \frac{37}{4}

Step 4 — the reciprocals. Put them over a common denominator:

\frac1\alpha + \frac1\beta = \frac{\beta + \alpha}{\alpha\beta} = \frac{7/2}{3/2} = \frac{7}{3}

Answers: \alpha^2+\beta^2 = \frac{37}{4} and \frac1\alpha+\frac1\beta = \frac73.

Verify by actually solving. 2x^2-7x+3 factorises as (2x-1)(x-3), so the roots are \tfrac12 and 3. Then \alpha^2+\beta^2 = 0.25 + 9 = 9.25 = \tfrac{37}{4} ✓, and 2 + \tfrac13 = \tfrac73 ✓.

What to notice. Any symmetric expression in the roots — one unchanged when you swap \alpha and \beta — can be written using only the sum and the product. That is why Vieta's formulas are more than a curiosity, and it is the starting point of the theory Galois built, told in 11.4.

Problem 5 — Simplifying an algebraic fraction

Simplify \dfrac{x^2-9}{x^2+x-12} \div \dfrac{x^2+6x+9}{x^2-16}.

Solution

Step 1 — turn the division into multiplication by flipping the second fraction.

\frac{x^2-9}{x^2+x-12} \times \frac{x^2-16}{x^2+6x+9}

Step 2 — factorise every single piece. Nothing can cancel until everything is in factors.

• x^2-9 = (x-3)(x+3) — difference of squares.
• x^2+x-12: two numbers multiplying to -12 and adding to +1 are 4 and -3, so (x+4)(x-3).
• x^2-16 = (x-4)(x+4) — difference of squares.
• x^2+6x+9 = (x+3)^2 — a perfect square, since 9 = 3^2 and 6 = 2\times3.

\frac{(x-3)(x+3)}{(x+4)(x-3)} \times \frac{(x-4)(x+4)}{(x+3)^2}

Step 3 — cancel matching factors from top and bottom. (x-3) goes, (x+4) goes, and one of the two (x+3) factors goes:

= \frac{x-4}{x+3}

Answer: \dfrac{x-4}{x+3}.

The part textbooks skip. This is valid only where the original expression exists. The starting expression is undefined at x = 3, x = -4, x = -3 and x = 4 (that last one because dividing by \frac{\ldots}{x^2-16} requires x^2-16 \ne 0). The simplified version looks fine at x = 3 and x = 4, but those points were never in the original's domain. Cancelling removes holes from the algebra without putting them back into the function.

Problem 6 — Solving a logarithmic equation, including the check

Solve \log_2(x) + \log_2(x-2) = 3.

Solution

Step 1 — combine the two logarithms into one with the product law \log b + \log c = \log(bc):

\log_2\big(x(x-2)\big) = 3

Step 2 — remove the logarithm using its definition. \log_2 A = 3 means A = 2^3:

x(x-2) = 8

Step 3 — solve the quadratic.

x^2 - 2x - 8 = 0

(x-4)(x+2) = 0 \quad \Rightarrow \quad x = 4 \text{ or } x = -2

Step 4 — check both in the original equation. This step is compulsory here, not optional.

• x = 4: gives \log_2 4 + \log_2 2 = 2 + 1 = 3 ✓
• x = -2: gives \log_2(-2), and there is no power of 2 that produces a negative number. This is not a valid solution.

Answer: x = 4 only.

What to notice. Step 1 was the step that lost information. Combining \log x + \log(x-2) into \log(x(x-2)) silently widens the domain, because x(x-2) is positive when both factors are negative, while the original needed each of x and x-2 to be positive on its own. The false root is a real solution of the transformed equation and a non-solution of the original — which is why every logarithmic equation ends with a substitution check.

Problem 7 — A binomial expansion and one specific term

Find the coefficient of x^5 in \left(2x^2 - \dfrac{3}{x}\right)^{7}.

Solution

Step 1 — write the general term. From the binomial theorem with a = 2x^2, b = -\frac3x, n = 7:

T_{k} = \binom{7}{k}(2x^2)^{7-k}\left(-\frac{3}{x}\right)^{k}

Step 2 — separate the powers of x from the numbers.

(2x^2)^{7-k} = 2^{7-k}x^{2(7-k)} = 2^{7-k}x^{14-2k}

\left(-\frac3x\right)^k = (-3)^k x^{-k}

So the power of x in the term is

x^{14-2k} \cdot x^{-k} = x^{14-3k}

Step 3 — find which k gives x^5.

14 - 3k = 5 \quad \Rightarrow \quad 3k = 9 \quad \Rightarrow \quad k = 3

k came out as a whole number between 0 and 7, so this term genuinely exists. (If it had come out fractional, the answer would be that there is no x^5 term at all — a common variation on this question.)

Step 4 — evaluate the coefficient at k=3.

\binom{7}{3} = \frac{7!}{3!\,4!} = \frac{7\times6\times5}{3\times2\times1} = 35

2^{7-3} = 2^4 = 16, \qquad (-3)^3 = -27

\text{coefficient} = 35 \times 16 \times (-27) = 560 \times (-27) = -15120

Answer: -15{,}120.

What to notice. You never expanded the whole thing. Writing the general term and solving for the exponent you want is the standard method, and it is the only sane approach for large n.

Problem 8 — Series in a real situation

You save £200 in month one and increase the amount by £15 every month after. Separately, a friend saves £200 in month one and increases by 4% each month. Who has saved more after 3 years, and by how much?

Solution

Yours is arithmetic: first term a_1 = 200, common difference d = 15, and n = 36 months.

Step 1 — use S_n = \frac{n}{2}\left(2a_1 + (n-1)d\right).

S_{36} = \frac{36}{2}\big(2(200) + 35(15)\big) = 18\big(400 + 525\big) = 18 \times 925 = £16{,}650

Sanity check with the other form. The last deposit is a_{36} = 200 + 35(15) = £725, and \frac{36}{2}(200 + 725) = 18 \times 925 ✓ — the average deposit is the average of the first and last, which is what Gauss's pairing proved.

Your friend's is geometric: a_1 = 200, ratio r = 1.04, n = 36.

Step 2 — use S_n = a_1\dfrac{1-r^n}{1-r}.

S_{36} = 200 \times \frac{1 - 1.04^{36}}{1 - 1.04}

Compute 1.04^{36} using logarithms: 36 \ln 1.04 = 36 \times 0.0392207 = 1.411947, so 1.04^{36} = e^{1.411947} = 4.10393.

S_{36} = 200 \times \frac{1 - 4.10393}{-0.04} = 200 \times \frac{-3.10393}{-0.04} = 200 \times 77.598 = £15{,}519.66

Answer: you are ahead, by 16{,}650 - 15{,}519.66 = £1{,}130.34.

What to notice — and it is the interesting part. The percentage saver is behind after three years but is gaining ground fast: their 36th deposit is 200 \times 1.04^{35} = £789 against your £725, and by month 48 they overtake you outright. Adding a fixed amount grows a total in a straight line; multiplying by a fixed factor grows it exponentially, and exponential growth always wins eventually, however slow it looks at the start. That single fact explains compound interest, epidemics, and why an algorithm's complexity class matters more than its constant factor.

Problem 9 — Partial fractions with a repeated factor

Split \dfrac{3x+5}{(x-1)^2(x+2)} into partial fractions.

Solution

Step 1 — write the right form. A repeated linear factor (x-1)^2 needs two terms, one for each power:

\frac{3x+5}{(x-1)^2(x+2)} = \frac{A}{x-1} + \frac{B}{(x-1)^2} + \frac{C}{x+2}

Leaving out the \frac{A}{x-1} term is the usual mistake, and it makes the system unsolvable.

Step 2 — multiply through by (x-1)^2(x+2) to clear every denominator:

3x+5 = A(x-1)(x+2) + B(x+2) + C(x-1)^2

Step 3 — choose values of x that kill terms.

Put x = 1: both the A and C terms contain (x-1), so both vanish.

3(1)+5 = B(1+2) \quad \Rightarrow \quad 8 = 3B \quad \Rightarrow \quad B = \frac83

Put x = -2: the A and B terms both contain (x+2) and vanish.

3(-2)+5 = C(-2-1)^2 \quad \Rightarrow \quad -1 = 9C \quad \Rightarrow \quad C = -\frac19

Step 4 — get A by comparing coefficients, since no value of x kills both B and C while leaving A. Look at the x^2 terms on each side. The left side has none, so its coefficient is 0. On the right, A(x-1)(x+2) contributes A x^2 and C(x-1)^2 contributes Cx^2:

0 = A + C \quad \Rightarrow \quad A = -C = \frac19

Step 5 — write the answer.

\frac{3x+5}{(x-1)^2(x+2)} = \frac{1/9}{x-1} + \frac{8/3}{(x-1)^2} - \frac{1/9}{x+2}

Check at a convenient value, x = 0. Left side: \frac{5}{(1)(2)} = 2.5. Right side: \frac{1/9}{-1} + \frac{8/3}{1} - \frac{1/9}{2} = -0.1111 + 2.6667 - 0.0556 = 2.5 ✓

Problem 10 — An inequality with a variable in the denominator

Solve \dfrac{x+1}{x-3} \le 2.

Solution

The wrong move first, because everyone makes it. Multiplying both sides by (x-3) looks natural and is invalid: x - 3 might be negative, and multiplying an inequality by a negative flips its direction. You do not know the sign in advance, so you cannot do it.

Step 1 — get everything on one side.

\frac{x+1}{x-3} - 2 \le 0

Step 2 — combine into a single fraction.

\frac{x+1 - 2(x-3)}{x-3} \le 0 \quad \Rightarrow \quad \frac{x + 1 - 2x + 6}{x-3} \le 0 \quad \Rightarrow \quad \frac{-x+7}{x-3} \le 0

Multiply top and bottom by -1 to make the leading sign positive. That flips the inequality once:

\frac{x-7}{x-3} \ge 0

Step 3 — find the points where the expression changes sign. A fraction can only change sign where the top is zero or the bottom is zero: at x = 7 and x = 3. These split the number line into three regions.

Step 4 — test one value in each region.

Region	Test	x-7	x-3	Fraction
x<3	x=0	-7	-3	positive ✓
3<x<7	x=5	-2	+2	negative ✗
x>7	x=8	+1	+5	positive ✓

Step 5 — handle the endpoints. At x=7 the fraction is 0, and the question allows \ge 0, so x=7 is included. At x = 3 the fraction is undefined — division by zero — so x=3 is excluded no matter what.

Answer: x < 3 or x \ge 7.

Check one from each part. x = 0: \frac{1}{-3} = -0.33 \le 2 ✓. x = 5: \frac{6}{2} = 3, which is not \le 2 ✓ correctly excluded. x = 8: \frac{9}{5} = 1.8 \le 2 ✓.

Problem 11 — A system of equations, two ways

Solve \begin{cases} 3x + 2y = 16 \\ 5x - 3y = 14 \end{cases}

Solution

Method 1 — elimination. Make the y coefficients cancel. Multiply the first equation by 3 and the second by 2, so both have |6y|:

9x + 6y = 48

10x - 6y = 28

Add them. The y terms are +6y and -6y, so they disappear:

19x = 76 \quad \Rightarrow \quad x = 4

Substitute back into the first original equation:

3(4) + 2y = 16 \quad \Rightarrow \quad 2y = 4 \quad \Rightarrow \quad y = 2

Method 2 — substitution. From the first equation, y = \frac{16-3x}{2}. Put that into the second:

5x - 3\left(\frac{16-3x}{2}\right) = 14

Multiply everything by 2 to clear the fraction:

10x - 3(16-3x) = 28 \quad \Rightarrow \quad 10x - 48 + 9x = 28 \quad \Rightarrow \quad 19x = 76 \quad \Rightarrow \quad x = 4

Same answer, as it must be.

Answer: x = 4, y = 2.

Check in both equations, not just one: 3(4)+2(2) = 16 ✓ and 5(4)-3(2) = 14 ✓.

What to notice. Elimination is faster when the coefficients are small whole numbers; substitution is better when one variable already sits alone. Both are special cases of the general elimination method of 4.3, which handles ten equations as comfortably as two.

Problem 12 — Complex arithmetic end to end

Let z = 3 + 4i and w = 1 - 2i. Find zw, \dfrac{z}{w}, |z| and the argument of z.

Solution

Step 1 — the product. Expand as normal brackets, then replace i^2 with -1:

zw = (3+4i)(1-2i) = 3 - 6i + 4i - 8i^2

Since i^2 = -1, the term -8i^2 becomes +8:

= 3 + 8 - 6i + 4i = 11 - 2i

Step 2 — the quotient. Multiply top and bottom by the conjugate of the bottom, \bar w = 1+2i:

\frac{z}{w} = \frac{(3+4i)(1+2i)}{(1-2i)(1+2i)}

Bottom first, using (a-bi)(a+bi) = a^2+b^2:

(1-2i)(1+2i) = 1 + 4 = 5

Top:

(3+4i)(1+2i) = 3 + 6i + 4i + 8i^2 = 3 - 8 + 10i = -5 + 10i

\frac{z}{w} = \frac{-5+10i}{5} = -1 + 2i

Step 3 — the modulus.

|z| = \sqrt{3^2+4^2} = \sqrt{25} = 5

Step 4 — the argument. z = 3+4i sits in the first quadrant, so

\arg z = \arctan\left(\frac43\right) = 0.9273 \text{ radians} = 53.13°

Answers: zw = 11-2i, \frac zw = -1+2i, |z| = 5, \arg z \approx 53.13°.

Check the quotient by multiplying back: (-1+2i)(1-2i) = -1 + 2i + 2i - 4i^2 = -1 + 4 + 4i = 3+4i = z ✓

Problem 13 — De Moivre generating an identity

Use De Moivre's theorem to derive a formula for \cos 3\theta in terms of \cos\theta alone.

Solution

Step 1 — write De Moivre with n = 3.

\cos3\theta + i\sin3\theta = (\cos\theta + i\sin\theta)^3

Step 2 — expand the right side with the binomial theorem, coefficients 1,3,3,1. Write c = \cos\theta, s = \sin\theta to keep it readable:

(c+is)^3 = c^3 + 3c^2(is) + 3c(is)^2 + (is)^3

Step 3 — simplify the powers of i. i^2 = -1 and i^3 = i^2 \cdot i = -i:

= c^3 + 3ic^2s - 3cs^2 - is^3

Step 4 — group real and imaginary parts.

= \underbrace{\left(c^3 - 3cs^2\right)}_{\text{real}} + i\underbrace{\left(3c^2s - s^3\right)}_{\text{imaginary}}

Step 5 — match parts. Two complex numbers are equal only when their real parts are equal and their imaginary parts are equal. Comparing with the left side:

\cos3\theta = c^3 - 3cs^2, \qquad \sin3\theta = 3c^2s - s^3

Step 6 — remove s from the cosine formula using s^2 = 1 - c^2:

\cos3\theta = c^3 - 3c(1-c^2) = c^3 - 3c + 3c^3 = 4c^3 - 3c

Answer: \cos3\theta = 4\cos^3\theta - 3\cos\theta.

Check at \theta = 60°, where \cos60° = 0.5 and \cos180° = -1:

4(0.125) - 3(0.5) = 0.5 - 1.5 = -1 \; ✓

What to notice. One line of complex algebra produced two trigonometric identities that would each take a page of geometry. This is the standard reason to reach for complex numbers: they turn statements about angles into statements about multiplication.

Problem 14 — A word problem that becomes a quadratic

A rectangular garden is 3 m longer than it is wide. A path 1 m wide is laid all the way around the outside, and the path uses 40 m² of paving. Find the garden's dimensions.

Solution

Step 1 — name the unknown and write both dimensions in terms of it. Let the width be w metres. Then the length is w+3.

Step 2 — write the outer rectangle's dimensions. The path adds 1 m on each side, so it adds 2 m to each measurement:

\text{outer width} = w+2, \qquad \text{outer length} = w+5

The factor of two here is the step people miss, and it is worth drawing to be sure.

Step 3 — the path's area is the difference of the two rectangles.

(w+2)(w+5) - w(w+3) = 40

Step 4 — expand both products.

\left(w^2 + 5w + 2w + 10\right) - \left(w^2 + 3w\right) = 40

w^2 + 7w + 10 - w^2 - 3w = 40

The w^2 terms cancel, which is a pleasant surprise — the equation is linear, not quadratic:

4w + 10 = 40 \quad \Rightarrow \quad 4w = 30 \quad \Rightarrow \quad w = 7.5

Answer: the garden is 7.5 m wide and 10.5 m long.

Check. Outer rectangle: 9.5 \times 12.5 = 118.75 m². Garden: 7.5 \times 10.5 = 78.75 m². Difference: 40 m² ✓

What to notice. The w^2 terms cancelling is not luck. The area of a border of width t around a rectangle p \times q is 2t(p+q) + 4t^2, which has no pq term in it at all — the border's area depends on the perimeter, not the area. That is a genuinely useful fact for anything involving edging, framing or coating.

Problem 15 — Everything at once

Solve 4^{x} - 5\cdot 2^{x} + 4 = 0.

Solution

Step 1 — spot the hidden structure. This looks like an exponential equation but it is a quadratic in disguise. The key observation is

4^x = (2^2)^x = (2^x)^2

using the power-of-a-power law. So if we call u = 2^x, then 4^x = u^2.

Step 2 — substitute.

u^2 - 5u + 4 = 0

Step 3 — solve the quadratic. Two numbers multiplying to 4 and adding to 5 are 1 and 4:

(u-1)(u-4) = 0 \quad \Rightarrow \quad u = 1 \text{ or } u = 4

Step 4 — undo the substitution. Remember u was never the answer; x is.

2^x = 1 \quad \Rightarrow \quad x = 0 \quad \text{(since } 2^0 = 1\text{)}

2^x = 4 \quad \Rightarrow \quad x = 2 \quad \text{(since } 2^2 = 4\text{)}

Step 5 — check for lost or false solutions. Here both values of u were positive, and 2^x can only ever be positive, so both survive. Had a root come out as u = -3, it would have been discarded on the spot, because no real x makes 2^x negative.

Answer: x = 0 and x = 2.

Verify. x=0: 1 - 5 + 4 = 0 ✓. x=2: 16 - 20 + 4 = 0 ✓.

What to notice. The move is called substitution, and the pattern to recognise is an expression and its square appearing in the same equation. It works on x^4 - 5x^2 + 4 = 0 (set u=x^2), on \sin^2\theta - 3\sin\theta + 2 = 0 (set u = \sin\theta), and on anything else with the same shape. Recognising a familiar problem wearing an unfamiliar costume is most of what mathematical skill actually is.

Next: Part 3 — Geometry, where the identities of this Part reappear as statements about shapes, and where 3.5 — trigonometry derives every trigonometric formula from a single circle.