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Reference Sheet: Programme F.6 — Polynomial 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. Polynomial equations

From Unit 01.

Evaluating a polynomial (F.3) takes a value of x and gives the value of the polynomial. Solving a polynomial equation reverses the process: set the polynomial equal to zero and find the values of x that satisfy it. On a graph, the solutions (roots) are where y=f(x) meets the x-axis.

2. Quadratic equations: the formula method

From Unit 01; derived in F.3 Unit 05.

ax^2+bx+c=0,\quad a\ne0:\qquad x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}.
  1. Read off a, b, c with their signs.
  2. Evaluate b^2-4ac and its square root to 4 dp.
  3. Evaluate both signs of \pm; round once, at the end, to 3 dp.

Examples.

5x^2+12x+3=0:\quad x=\frac{-12\pm\sqrt{144-60}}{10}=\frac{-12\pm9.1652}{10}=-0.283\ \text{or}\ -2.117
3x^2-10x+4=0:\quad x=\frac{10\pm\sqrt{100-48}}{6}=\frac{10\pm7.2111}{6}=2.869\ \text{or}\ 0.465
x^2+15x-7=0:\quad x=\frac{-15\pm\sqrt{225+28}}{2}=\frac{-15\pm15.9060}{2}=0.453\ \text{or}\ -15.453

3. Beyond the book: reading the formula as centre and half-gap

From Unit 01.

x=\underbrace{-\frac b{2a}}_{\text{centre}}\pm\underbrace{\frac{\sqrt{b^2-4ac}}{2a}}_{\text{half-gap}}

The two roots sit symmetrically about the parabola's axis of symmetry x=-\frac b{2a}, which passes through its turning point.

Example. 6x^2-8x-9=0: centre \frac8{12}=0.6667, half-gap \frac{\sqrt{280}}{12}=1.3944, so x=2.061 or -0.728.

4. Beyond the book: how many real roots

From Unit 01.

b^2-4ac Real roots Parabola and axis
>0 two different meets it twice, crossing
=0 one repeated touches it
<0 none (two complex roots) never reaches it

Examples. The stone h=4+12t-5t^2: reaching 12 m needs 5t^2-12t+8=0, with b^2-4ac=-16 — never; reaching 11.2 m needs 5t^2-12t+7.2=0, with b^2-4ac=0 — once, at the peak, t=1.2 s. The book's quadratics all have b^2-4ac>0.

5. Cubic equations: nested form

From Unit 02; nesting is F.3 Unit 03.

Sections 5–8 are the book's cubic equations having at least one simple linear factor: the method needs one factor (x-k) to be found by trial, and fails without it (§13).

Write the cubic in descending powers, with a zero coefficient for any missing power, and nest it so each trial is one chain of multiply-and-add.

Examples.

2x^3-11x^2+18x-8=[(2x-11)x+18]x-8
x^3-5x-2=[(x+0)x-5]x-2

6. Cubic equations: finding a simple linear factor

From Unit 02.

Apply the remainder theorem: evaluate f(1),f(-1),f(2),f(-2),\ldots in that order until f(k)=0. Record each failure ("(x-1) is not a factor"). Then (x-k) is a factor; f(-2)=0 means (x+2).

Example. f(x)=5x^3+2x^2-26x-20=[(5x+2)x-26]x-20:

trial 1 -1 2 -2
f -39 3 -24 0

So (x+2) is a factor.

7. Cubic equations: the quadratic factor by long division

From Unit 02; the layout is in F.2 Unit 05.

Divide f(x) by (x-k) to get f(x)=(x-k)(ax^2+bx+c). The remainder must be 0 — if it is not, the trial was wrong.

Example.

\begin{array}{r|rrrr} & x^3 & x^2 & x^1 & x^0 \\ \hline \text{dividend} & 5 & 2 & -26 & -20 \\ \text{subtract } 5x^2(x+2) & 5 & 10 & & \\ \hline & & -8 & -26 & -20 \\ \text{subtract } -8x(x+2) & & -8 & -16 & \\ \hline & & & -10 & -20 \\ \text{subtract } -10(x+2) & & & -10 & -20 \\ \hline \text{remainder} & & & & 0 \end{array}

The quotient is 5x^2-8x-10.

8. Cubic equations: solving the quadratic factor

From Unit 02.

(x-k)(ax^2+bx+c)=0 means x=k or ax^2+bx+c=0, because a product is zero only when a factor is. Solve the quadratic by the formula (§2). Three solutions: x=k, x=x_1, x=x_2.

Examples.

5x^3+2x^2-26x-20=0:\quad 5x^2-8x-10=0,\ x=\frac{8\pm16.2481}{10};\quad x=-2,\ 2.425,\ -0.825
2x^3-11x^2+18x-8=0:\quad f(2)=0,\ 2x^2-7x+4=0,\ x=\frac{7\pm4.1231}4;\quad x=2,\ 2.781,\ 0.719

9. Beyond the book: roots the situation rules out

From Unit 01 and Unit 02.

The equation gives every root; the physical situation decides which apply. Apply the constraint after solving, and keep every root that survives it — often more than one does.

Examples. The stone lands at t=2.697 s; t=-0.297 s is before it was thrown. The 12\times8 cm tray holding 64 cm^3 needs 0<x<4: x=2 and x=1.172 both work, x=6.828 does not.

10. Fourth-order equations: the sequential route

From Unit 03.

The method needs at least two simple linear factors.

  1. Find a first factor (x-k) by trial in nested form (§6).
  2. Divide: f(x)=(x-k)\,F(x), with F(x) a cubic.
  3. f(x)=0 gives x=k or F(x)=0.
  4. Re-nest F(x) and solve it as a cubic (§6–§8), starting the trials again at x=1.
  5. Four solutions: x=k, x=m, x=x_1, x=x_2.

Example. 2x^4-4x^3-23x^2-11x+6=0. f(1)=-30, f(-1)=0: f(x)=(x+1)(2x^3-6x^2-17x+6). Then F(1)=-15, F(-1)=15, F(2)=-36, F(-2)=0: F(x)=(x+2)(2x^2-10x+3).

x=\frac{10\pm\sqrt{76}}4=\frac{10\pm8.7178}4;\qquad x=-1,\ -2,\ 4.679,\ 0.321

11. Repeated roots

From Unit 03.

A factor can divide more than once. Always retry a root already found. A repeated root is listed as often as its factor occurs; the graph touches the axis there instead of crossing it.

Example. 3x^4+2x^3-15x^2+12x-2=0: f(1)=0, f(x)=(x-1)(3x^3+5x^2-10x+2); F(1)=0 again, so f(x)=(x-1)^2(3x^2+8x-2).

x=1,\quad x=1,\quad x=\frac{-8\pm9.3808}6=0.230,\ -2.897

12. Two factors at once (route 2)

From Unit 03; introduced in F.3 Unit 06.

If the first hunt finds two roots p and q, multiply (x-p)(x-q) and divide once. The quadratic factor is the same as by the sequential route.

Example. 2x^4+3x^3-13x^2-6x+8=0: f(1)=-6, f(-1)=0, f(2)=0. (x+1)(x-2)=x^2-x-2, and

\frac{2x^4+3x^3-13x^2-6x+8}{x^2-x-2}=2x^2+5x-4;\qquad x=-1,\ 2,\ \frac{-5\pm7.5498}4=0.637,\ -3.137

13. The limits of the method

From Unit 04.

Everything above relies on finding simple factors by the remainder theorem: at least one for a cubic, at least two for a fourth-order equation. Many cubic and fourth-order equations have no simple factors. Their real roots still exist wherever the graph meets the axis, but other (numerical) methods are needed, later in the course.

Example. The floating ball, x^3-3x^2+3=0: f(1)=1, f(-1)=-1, f(2)=-1, f(-2)=-17, f(3)=3, f(-3)=-51. No trial works; the roots are -0.879, 1.347, 2.532.

14. Beyond the book: locating a root by a change of sign

From Unit 04.

If f(p) and f(q) have opposite signs, the graph crosses the axis between p and q. The trial table already shows where to look; narrowing the interval is the start of a numerical method. To quote a root to 3 dp, bracket it between the two half-way values that round to it. The converse fails: equal signs at p and q do not rule out roots between them, since two can hide between the samples.

Example. x^3-3x^2+3: f(1)=1, f(2)=-1; f(1.3)=0.127, f(1.4)=-0.136; f(1.34)=0.019, f(1.35)=-0.007; f(1.3465)=0.0021, f(1.3475)=-0.0005. So x=1.347 (3 dp).

Example of the converse. 2x^3-20x^2+48x-33: f(1)=-3 and f(2)=-1 have the same sign, but f(1.5)=0.75, and there are roots near 1.30 and 1.85.

15. Beyond the book: candidates and checks

From Unit 04 and Unit 01.

Rational candidates (the rational root theorem, proved in Unit 04). With whole-number coefficients, a root \frac pq in lowest terms has p dividing the constant term and q dividing the leading coefficient. If every candidate fails, there is no simple linear factor. x^3-3x^2+3: candidates \pm1,\pm3, all fail.

Root checks. For ax^2+bx+c=0 (roots m\pm d, Unit 01): x_1+x_2=-\frac ba and x_1x_2=m^2-d^2=\frac ca. For a cubic ax^3+bx^2+cx+d=0 with roots p,q,r, expanding a(x-p)(x-q)(x-r) (Unit 04) shows that they add up to -\frac ba.

Examples. 2x^2-3x-4=0: 2.351+(-0.851)=1.5=\frac32, and \frac{3+\sqrt{41}}4\times\frac{3-\sqrt{41}}4=\frac{9-41}{16}=-2=\frac{-4}{2}. 2x^3-11x^2+18x-8=0: 2+0.719+2.781=5.5=\frac{11}2.

Traps

  • Dropping a sign in the formula. For 2x^2-3x-4, b=-3: -b=+3 and b^2=(-3)^2=+9. Writing -3 for -b gives roots of the wrong sign.
  • Rounding early. Keep \sqrt{41}=6.4031 to 4 dp and round the final answers; rounding \sqrt{41} to 6.4 gives -0.850 instead of -0.851.
  • Reporting one root. The \pm gives two roots. Over \mathbb C, counted with multiplicity, a cubic has three roots and a fourth-order equation four; list every one the method finds, a repeated root as often as its factor occurs.
  • Trials out of order. Use 1,-1,2,-2,\ldots: the negatives count. In 3x^3+12x^2+13x+4 the first factor is (x+1), and trying only positive values never finds it.
  • The wrong sign of the factor. f(-2)=0 gives (x+2), not (x-2).
  • Missing zero coefficient. Nesting x^3-5x-2 as (x-5)x-2 drops a power of x; it is [(x+0)x-5]x-2. Long division needs the same placeholder.
  • Not retrying a root. After f(1)=0, try F(1) again: in 3x^4+2x^3-15x^2+12x-2 it is zero, and skipping it leaves no factor to find.
  • Dropping the leading coefficient. The quadratic factor keeps it: 3x^3+12x^2+13x+4=(x+1)(3x^2+9x+4), not (x+1)(x^2+9x+4).
  • Rejecting a valid root. Apply the physical constraint, but keep every root that satisfies it: the 64 cm^3 tray has two valid cuts.
  • "No trial works, so no roots." x^3-3x^2+3=0 has three real roots; they are just not simple numbers. And b^2-4ac<0 means no real roots — there are two complex ones.

Self-check

  1. Solve 7x^2+3x-2=0, to 3 dp.
  2. The stone h=4+12t-5t^2 (metres, seconds): when is it 8 m up?
  3. Show that the same stone never reaches 12 m.
  4. Solve x^3-4x^2+x+2=0.
  5. Solve 2x^3+x^2-7x-2=0.
  6. Solve x^3-3x^2-2x+2=0.
  7. Squares of side x cm are cut from the corners of a 10 cm by 6 cm card to fold a tray of volume 32 cm^3. Find every possible cut.
  8. Solve x^4-x^3-5x^2+3x+2=0.
  9. Solve 3x^4-4x^3-6x^2+3x+2=0.
  10. Solve x^4-2x^3-6x^2-2x+1=0, and say what its graph does at the repeated root.
  11. Beyond the book: show that x^3+x^2-2x-1=0 has no simple linear factor, and find an interval of width 0.1 containing its positive root.
  12. A fourth-order polynomial with leading coefficient 1 touches the x-axis at x=3 and crosses it at x=-1 and x=2. Write it out, and check it with the first two trials.

Fully worked solutions

1. a=7, b=3, c=-2: x=\frac{-3\pm\sqrt{9+56}}{14}=\frac{-3\pm8.0623}{14}, so x=0.362 or x=-0.790.

2. 4+12t-5t^2=8 gives 5t^2-12t+4=0: t=\frac{12\pm\sqrt{144-80}}{10}=\frac{12\pm8}{10}, so t=0.4 s (rising) and t=2 s (falling). Both are valid, and they sit symmetrically about the peak at 1.2 s.

3. 4+12t-5t^2=12 gives 5t^2-12t+8=0, with b^2-4ac=144-160=-16<0: no real t. The peak height is h(1.2)=4+14.4-7.2=11.2 m, short of 12.

4. Nested [(x-4)x+1]x+2; f(1): 1\to-3\to-2\to0, so (x-1) is a factor and f(x)=(x-1)(x^2-3x-2). Then x=\frac{3\pm\sqrt{9+8}}2=\frac{3\pm4.1231}2. So x=1, 3.562, -0.562.

5. Nested [(2x+1)x-7]x-2: f(1)=-6, f(-1)=4, f(2)=4, f(-2): 2\to-3\to-1\to0. So f(x)=(x+2)(2x^2-3x-1), and x=\frac{3\pm\sqrt{9+8}}4=\frac{3\pm4.1231}4. So x=-2, 1.781, -0.281.

6. Nested [(x-3)x-2]x+2: f(1)=-2; f(-1): 1\to-4\to2\to0. So f(x)=(x+1)(x^2-4x+2), and x=\frac{4\pm\sqrt{16-8}}2=2\pm1.4142. So x=-1, 3.414, 0.586.

7. x(10-2x)(6-2x)=32 gives 4x^3-32x^2+60x-32=0, i.e. x^3-8x^2+15x-8=0. f(1): 1\to-7\to8\to0, so f(x)=(x-1)(x^2-7x+8), and x=\frac{7\pm\sqrt{49-32}}2=\frac{7\pm4.1231}2=5.562 or 1.438. The 6 cm side needs 0<x<3, so the cuts are x=1 cm (8\times4\times1 tray) and x=1.438 cm; 5.562 is rejected.

8. f(1): 1\to0\to-5\to-2\to0, so f(x)=(x-1)(x^3+0x^2-5x-2). For F(x)=[(x+0)x-5]x-2: F(1)=-6, F(-1)=2, F(2)=-4, F(-2)=0, so F(x)=(x+2)(x^2-2x-1) and x=1\pm\sqrt2. So x=1, -2, 2.414, -0.414.

9. f(1)=-2; f(-1): 3\to-7\to1\to2\to0, so f(x)=(x+1)(3x^3-7x^2+x+2). For F: F(1)=-1, F(-1)=-9, F(2): 3\to-1\to-1\to0, so F(x)=(x-2)(3x^2-x-1) and x=\frac{1\pm\sqrt{1+12}}6=\frac{1\pm3.6056}6. So x=-1, 2, 0.768, -0.434.

10. f(1)=-8; f(-1): 1\to-3\to-3\to1\to0, so f(x)=(x+1)(x^3-3x^2-3x+1). For F: F(1)=-4, F(-1): 1\to-4\to1\to0, so F(x)=(x+1)(x^2-4x+1) and x=2\pm\sqrt3. So x=-1, -1, 3.732, 0.268. At x=-1 the factor (x+1)^2 does not change sign, so the graph touches the axis there.

11. f(1)=-1, f(-1)=1, f(2)=7, f(-2)=-1. Beyond the book: by the rational root theorem (§15) the only rational candidates are \pm1, since the constant term and leading coefficient are \pm1; both fail, so there is no simple linear factor. f(1)<0<f(2), and f(1.2)=-0.232, f(1.3)=0.287, so the positive root lies between 1.2 and 1.3 (it is 1.247).

12. Touching at 3 means (x-3)^2; crossing at -1 and 2 means (x+1) and (x-2): $f(x)=(x-3)^2(x+1)(x-2)=x^4-7x^3+13x^2+3x-18.Trials:f(1)=1-7+13+3-18=-8, so(x-1)is not a factor;f(-1)=1+7+13-3-18=0, so(x+1)$ is, as built.