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Reference Sheet: Programme F.3 — Expressions and Equations

Rules, procedures and drill, in the order of the book's programme. Every rule links to the unit in the lesson that derives it — if a rule here looks arbitrary, the link is where it stops being arbitrary.

1. Evaluating expressions

From Unit 01.

Substitute numbers for the letters and follow the order of operations. Give the answer to the stated number of decimal places (dp) or significant figures (sf), using the stated value of \pi (3.142, 3.14 or 3.14159\ldots). Round once, at the end.

Examples.

2\pi\sqrt{\frac lg}\ \text{at}\ l=2,\ g=9.81:\quad 2.84\ \text{(2 dp)}
R=\frac{R_1R_2}{R_1+R_2}\ \text{at}\ R_1=276,\ R_2=145:\quad R=\frac{40\,020}{421}=95.06
V=\frac{\pi h}6(3R^2+h^2)\ \text{at}\ h=2.85,\ R=6.24,\ \pi=3.142:\quad V=186.46

2. The five types of equation

From Unit 01.

Type Meaning Symbol Example
Conditional equation true only for certain values = x^2=4 (true for x=2, x=-2)
Identity true for all values where both sides are defined \equiv 2(5-x)\equiv10-2x
Defining equation fixes what notation means a "defined as" sign a^2 means a\times a
Assigning equation gives a variable a value := p:=4
Formula states a fact linking well-defined quantities = A=\pi r^2

Writing = for all five is common and acceptable. An identity with a denominator holds only where both sides are defined: \frac{r^3-s^3}{r-s}\equiv r^2+rs+s^2 for r\ne s.

3. Subject, dependent and independent variables

From Unit 01.

In r=2s^3+3t: r is the subject and the dependent variable; s and t are the independent variables.

4. Evaluating an independent variable

From Unit 02.

Given the subject's value and all but one independent variable, undo the operations on the missing one, doing the same to both sides.

Examples.

T=2\pi\sqrt{\frac lg},\ T=1.03,\ \pi=3.14,\ g=9.81:\quad l=9.81\left(\frac{1.03}{6.28}\right)^2=0.264
I=\frac{nE}{R+nr},\ n=6,\ E=2.01,\ R=12,\ I=0.98:\quad r=\frac16\left(\frac{12.06}{0.98}-12\right)=0.051

5. Transposition of formulas

From Unit 02.

To change the subject: locate the target, then remove the operations around it in reverse order, each by its opposite — addition ↔ subtraction, multiplication ↔ division, powers ↔ roots (and reciprocals of both sides). Do everything to both sides. Write the new subject on the left.

Examples.

T=2\pi\sqrt{\frac lg}\;\Rightarrow\;l=\frac{gT^2}{4\pi^2}
v=u+at\;\Rightarrow\;a=\frac{v-u}t
a=\frac{2(ut-s)}{t^2}\;\Rightarrow\;at^2+2s=2ut\;\Rightarrow\;u=\frac{at^2+2s}{2t}
d=2\sqrt{h(2r-h)}\;\Rightarrow\;\frac{d^2}{4h}=2r-h\;\Rightarrow\;r=\frac{d^2+4h^2}{8h}
V=\frac{\pi h(3R^2+h^2)}6\;\Rightarrow\;R=\sqrt{\frac{2V}{\pi h}-\frac{h^2}3}

6. When the target appears twice

From Unit 02.

Clear fractions, multiply out, collect every term containing the target on one side, and factor it out.

Examples.

n=\frac{IR}{E-Ir}\;\Rightarrow\;nE=I(R+nr)\;\Rightarrow\;I=\frac{nE}{R+nr}
\frac Rr=\sqrt{\frac{f+P}{f-P}}\;\Rightarrow\;R^2(f-P)=r^2(f+P)\;\Rightarrow\;f=\frac{(R^2+r^2)P}{R^2-r^2}

7. The evaluation process and f(x)

From Unit 02.

A system takes an input, processes it and gives an output. With input x and process f, the output is f(x), "f acting on x".

Examples. f(x)=3x-4: f(5)=11. f(x)=4x^3-\frac6{2x}: f(3)=108-1=107; f(-4)=-256+0.75=-255.25.

8. Polynomial expressions

From Unit 03.

A polynomial in x is a sum of terms in powers of x, written in descending powers. The degree is the highest power: linear (1), quadratic (2), cubic (3), quartic or fourth-order (4). 5x^4+7x^3+3x-4 has degree 4.

9. Evaluation of a polynomial by nesting

From Unit 03.

  1. Write the terms in descending powers.
  2. Insert every missing power with a zero coefficient.
  3. Starting with the leading coefficient, repeatedly multiply by x and add the next coefficient.

Examples.

5x^3+2x^2-3x+6=[(5x+2)x-3]x+6;\quad f(4):\ 5\to22\to85\to346
3x^4+2x^2-4x+5=\{[(3x+0)x+2]x-4\}x+5;\quad f(2):\ 3\to6\to14\to24\to53
x^4-3x^3+2x-3=\{[(x-3)x+0]x+2\}x-3;\quad f(5)=257

10. Remainder theorem

From Unit 04.

f(x)=(x-a)\,g(x)+R,\qquad \frac{f(x)}{x-a}=g(x)+\frac R{x-a},\qquad R=f(a).

For a divisor x+k, use a=-k.

Examples. (x^3+3x^2-13x-10)\div(x-3): R=f(3)=5. (5x^3-4x^2-3x+6)\div(x-2): R=24. (2x^3+3x^2-x+4)\div(x+2): R=f(-2)=2.

11. Factor theorem

From Unit 04.

If f(a)=0 then (x-a) is a factor of f(x), and conversely. A sketch suggests candidates for a; only f(a)=0 exactly confirms one.

Example. x^3-5x^2-2x+24=[(x-5)x-2]x+24; f(3)=0, so (x-3) is a factor.

12. Finding the other factor: long division

From Unit 04; the layout is in F.2 Unit 05.

Divide f(x) by (x-a) with the long-division layout, inserting zero coefficients for missing powers.

Example. (x^3+3x^2-13x-10)\div(x-3)=x^2+6x+5, remainder 5.

Alternative (synthetic division, beyond the book). In the nesting chain for f(a), the values before the last are the quotient's coefficients and the last is the remainder: 1\to6\to5\to5 gives x^2+6x+5, remainder 5. Use it to check a long division, not instead of learning one.

13. Finding a first factor by trial

From Unit 06.

Write f(x) in nested form and try x=1,-1,2,-2,3,\ldots until f(k)=0. Record each failed trial.

Example. x^3+5x^2-2x-24: f(1)=-20, f(-1)=-18, f(2)=0, so (x-2) is a factor.

14. The general quadratic equation

From Unit 05.

For ax^2+bx+c=0, a\ne0, complete the square by adding \left(\frac b{2a}\right)^2 — the square of half the coefficient of x after dividing by a — to both sides:

x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}.

Examples. 2x^2+5x-3=0: x=\frac{-5\pm7}4=\frac12,\ -3. 3x^2-4x-2=0: x=\frac{4\pm\sqrt{40}}6=1.721,\ -0.387 (3 dp).

15. From roots to factors

From Unit 05.

ax^2+bx+c=a(x-r_1)(x-r_2).

Example. Roots \frac12 and -3: 2x^2+5x-3=2\left(x-\frac12\right)(x+3)=(2x-1)(x+3).

16. Factorizing a cubic

From Unit 06.

Trial for one factor (x-k); divide to leave a quadratic; factorize the quadratic, by the formula if necessary.

Examples.

2x^3+x^2-13x+6:\ f(2)=0\;\Rightarrow\;(x-2)(2x^2+5x-3)=(x-2)(x+3)(2x-1)
x^3-6x^2-7x+60:\ f(4)=0\;\Rightarrow\;(x-4)(x^2-2x-15)=(x-4)(x-5)(x+3)
2x^3-9x^2+7x+6:\ f(2)=0\;\Rightarrow\;(x-2)(2x^2-5x-3)=(x-2)(x-3)(2x+1)

17. Fourth-order polynomials, route 1

From Unit 06.

Find one factor, divide to leave a cubic g(x), then factorize g(x) as in §16. The method needs at least one simple factor.

Example. 2x^4-x^3-8x^2+x+6: f(1)=0 gives (x-1)(2x^3+x^2-7x-6); g(1)=-10, g(-1)=0 gives (x+1)(2x^2-x-6); so (x-1)(x+1)(x-2)(2x+3).

18. Fourth-order polynomials, route 2

From Unit 06.

Find two factors, multiply them into a quadratic, divide once: f(x)=(x-p)(x-q)(ax^2+bx+c).

Example. 2x^4-5x^3-15x^2+10x+8: f(1)=0, f(-1)=-10, f(2)=-40, f(-2)=0; divide by (x-1)(x+2)=x^2+x-2 to get 2x^2-7x-4=(x-4)(2x+1); so (x-1)(x+2)(x-4)(2x+1).

Traps

  • \equiv on a formula. A=\pi r^2 is a formula, not an identity; \equiv means "true for every value of the variables".
  • An identity outside its domain. \frac{r^3-s^3}{r-s}\equiv r^2+rs+s^2 says nothing at r=s, where the left side is undefined.
  • Undoing an inner operation first. In T=2\pi\sqrt{l/g} the division by g happens inside the root, so it is undone last. Multiplying both sides by g first gives gT=2\pi g\sqrt{l/g}, which frees nothing. Undo the outermost operation (\times2\pi) first.
  • Dividing by something that may be zero. Collecting the target and dividing by its factor, e.g. (R+nr), assumes that factor is not zero.
  • Checking a rearrangement in the rearranged formula. Check it in the original: the rearranged one inherits any mistake.
  • Nesting without a placeholder. 3x^4+2x^2-4x+5 at x=2 without the 0x^3 gives 29 instead of 53.
  • The sign of a. The remainder on dividing by x+2 is f(-2), not f(2).
  • Forgetting the leading coefficient. Roots \frac12 and -3 give 2x^2+5x-3=2(x-\frac12)(x+3), not (x-\frac12)(x+3).
  • Half the coefficient. Completing x^2+\frac52x needs \left(\frac54\right)^2, the square of half the coefficient.
  • Reading factors off a sketch. A sketch gives candidates; only f(a)=0 gives a factor.

Self-check

  1. Classify: (a) x^2-5x+6=0; (b) (x+1)^2\equiv x^2+2x+1; (c) k:=9.81; (d) E=mc^2; (e) 3y means 3\times y.
  2. Evaluate V=\frac{\pi h}6(3R^2+h^2) for h=1.20, R=2.50, \pi=3.142, to 3 dp.
  3. Make a the subject of s=ut+\frac12at^2.
  4. Make u the subject of v^2=u^2+2as.
  5. Make x the subject of y=\frac{x+2}{x-3}.
  6. A circuit resonates at f=\frac1{2\pi\sqrt{LC}}. Find the capacitance C that tunes a 0.2 H coil to f=50 Hz, to 3 sf.
  7. Write 3x^4-2x^3+5x-7 in nested form and find f(-2).
  8. Find the remainder when 2x^3-5x^2+4x-9 is divided by x+1.
  9. Show that (x-2) is a factor of x^3-3x^2-10x+24 and factorize completely.
  10. Solve 3x^2-4x-2=0 to 3 dp.
  11. Factorize 3x^3+4x^2-5x-2.
  12. Factorize x^4-x^3-7x^2+x+6 into four linear factors.

Fully worked solutions

1. (a) Conditional equation: true only for x=2 and x=3. (b) Identity: true for every x. (c) Assigning equation. (d) Formula. (e) Defining equation: it fixes the notation.

2. 3R^2+h^2=18.75+1.44=20.19; \frac{3.142\times1.20}6=0.6284; V=0.6284\times20.19=12.687.

3. s-ut=\frac12at^2, so 2(s-ut)=at^2 and a=\frac{2(s-ut)}{t^2}.

4. u^2=v^2-2as, so u=\sqrt{v^2-2as} (the positive root, if u is a speed).

5. y(x-3)=x+2, so xy-3y=x+2; collect: xy-x=3y+2; factor: x(y-1)=3y+2; so x=\frac{3y+2}{y-1}, for y\ne1.

6. 2\pi\sqrt{LC}=\frac1f, so LC=\frac1{4\pi^2f^2} and C=\frac1{4\pi^2f^2L}=\frac1{4\pi^2\times2500\times0.2}=\frac1{19\,739}=5.07\times10^{-5} F, about 50.7\ \muF.

7. Insert 0x^2: \{[(3x-2)x+0]x+5\}x-7. At x=-2: 3\to-8\to16\to-27\to47. So f(-2)=47.

8. a=-1: f(-1)=-2-5-4-9=-20.

9. Nested [(x-3)x-10]x+24 at x=2: 1\to-1\to-12\to0, so (x-2) is a factor and the quotient is x^2-x-12=(x-4)(x+3). So x^3-3x^2-10x+24=(x-2)(x-4)(x+3).

10. a=3, b=-4, c=-2: x=\frac{4\pm\sqrt{16+24}}6=\frac{4\pm6.3246}6=1.721 or -0.387.

11. f(1)=3+4-5-2=0, so (x-1) is a factor. Dividing: 3x^2+7x+2=(3x+1)(x+2). So (x-1)(3x+1)(x+2).

12. f(1)=1-1-7+1+6=0 and f(-1)=1+1-7-1+6=0, so (x-1)(x+1)=x^2-1 divides f. Dividing gives x^2-x-6=(x-3)(x+2). So (x-1)(x+1)(x-3)(x+2).