A review set: evaluating expressions with rounding, making a variable the subject inside a double root, the nested form, remainders, and factoring polynomials into linear factors.

This is the last lesson of the algebra part and at the same time a test of it. The exercises are deliberately mixed: for each one you first have to recognise which technique is wanted and only then compute. If any of them gives you trouble, each section says which lesson it comes from.

Evaluating expressions

Calculate correct to three significant figures.

\sqrt{\frac{2V}{\pi h} - \frac{h^{2}}{3}}

\text{a)}\quad V = 23.05

h = 2.69

\pi = 3.14

\begin{aligned}\sqrt{\frac{2V}{\pi h} - \frac{h^{2}}{3}} &= \sqrt{\frac{2\cdot 23.05}{3.14\cdot 2.69} - \frac{2.69^{2}}{3}} = \sqrt{\frac{46.1}{8.45} - \frac{7.24}{3}} = \\&= \sqrt{5.46 - 2.41} = \sqrt{3.05} = 1.75\end{aligned}

\text{b)}\quad V = 85.67

h = 5.44

\pi = 3.14

\begin{aligned}\sqrt{\frac{2V}{\pi h} - \frac{h^{2}}{3}} &= \sqrt{\frac{2\cdot 85.67}{3.14\cdot 5.44} - \frac{5.44^{2}}{3}} = \sqrt{\frac{171.34}{17.1} - \frac{29.6}{3}} = \\&= \sqrt{10.02 - 9.87} = \sqrt{0.15} = 0.387\end{aligned}

Rearranging a formula

Given q = √((5p² − 1)² − 2), express p in terms of q.

\begin{aligned}q &= \sqrt{(5p^{2} - 1)^{2} - 2} \;\Big/ ^{2} \\q^{2} &= (5p^{2} - 1)^{2} - 2 \\q^{2} + 2 &= (5p^{2} - 1)^{2} \;\Big/ \sqrt{\phantom{x}} \\\sqrt{q^{2} + 2} &= 5p^{2} - 1 \\\sqrt{q^{2} + 2} + 1 &= 5p^{2} \;\Big/ : 5 \\\frac{\sqrt{q^{2} + 2} + 1}{5} &= p^{2} \;\Big/ \sqrt{\phantom{x}} \\p &= \sqrt{\frac{\sqrt{q^{2} + 2} + 1}{5}}\end{aligned}

Horner's method

Express each polynomial in nested form and determine its value for the given x.

\text{a)}\quad f(x) = 2x^{3} - 3x^{2} + 5x - 4 = [(2x - 3)x + 5]x - 4

f(3) = [(2\cdot 3 - 3)\cdot 3 + 5]\cdot 3 - 4 = 38

\text{b)}\quad f(x) = 4x^{3} + 2x^{2} - 7x - 2 = [(4x + 2)x - 7]x - 2

f(2) = [(4\cdot 2 + 2)\cdot 2 - 7]\cdot 2 - 2 = 24

\text{c)}\quad f(x) = 3x^{3} - 4x^{2} + 8 = 3x^{3} - 4x^{2} + 0x + 8 = [(3x - 4)x + 0]x + 8

f(4) = [(3\cdot 4 - 4)\cdot 4 + 0]\cdot 4 + 8 = 136

\text{d)}\quad f(x) = 2x^{3} + x - 5 = 2x^{3} + 0x^{2} + x - 5 = [(2x + 0)x + 1]x - 5

f(5) = [(2\cdot 5 + 0)\cdot 5 + 1]\cdot 5 - 5 = 250

The remainder

Determine the remainder obtained from the division.

(4x^{3} - 5x^{2} + 7x - 3) : (x - 3)

[(4x - 5)x + 7]x - 3

f(3) = [(4\cdot 3 - 5)\cdot 3 + 7]\cdot 3 - 3 = 81

this is the remainder

Factoring

Check whether x − 2 is a factor of the polynomial f(x). If so, factorise f(x).

f(x) = 2x^{3} + 2x^{2} - 17x + 10 = [(2x + 2)x - 17]x + 10

f(2) = [(2\cdot 2 + 2)\cdot 2 - 17]\cdot 2 + 10 = 0

(2x^{3} + 2x^{2} - 17x + 10) : (x - 2) = 2x^{2} + 6x - 5

$2x^{3}$	$- 4x^{2}$
0	$6x^{2}$	$- 17x$
	$6x^{2}$	$- 12x$
	0	$- 5x$	$+ 10$
		$- 5x$	$+ 10$
			$0$

(x - 2)(2x^{2} + 6x - 5) = 2x^{3} + 2x^{2} - 17x + 10

2x^{3} + 2x^{2} - 17x + 10 = (x - 2)(2x^{2} + 6x - 5)

Write f(x) = 2x³ + 7x² − 14x − 40 as a product of three linear factors.

f(x) = 2x^{3} + 7x^{2} - 14x - 40 = [(2x + 7)x - 14]x - 40

x = -2 \;\Rightarrow\; f(-2) = [(2\cdot(-2) + 7)\cdot(-2) - 14]\cdot(-2) - 40 = 0

(2x^{3} + 7x^{2} - 14x - 40) : (x + 2) = 2x^{2} + 3x - 20

$2x^{3}$	$+ 4x^{2}$
0	$3x^{2}$	$- 14x$
	$3x^{2}$	$+ 6x$
	0	$- 20x$	$- 40$
		$- 20x$	$- 40$
			$0$

(x + 2)(2x^{2} + 3x - 20) = 2x^{3} + 7x^{2} - 14x - 40

x = \frac{-3 \pm \sqrt{3^{2} - 4\cdot 2\cdot(-20)}}{2\cdot 2} = \frac{-3 \pm \sqrt{9 + 160}}{4} = \frac{-3 \pm 13}{4}

x_{1} = -4

x_{2} = \tfrac{5}{2}

2x^{2} + 3x - 20 = 2(x + 4)\left(x - \tfrac{5}{2}\right) = (x + 4)(2x - 5)

2x^{3} + 7x^{2} - 14x - 40 = (x + 2)(x + 4)(2x - 5)

Express f(x) = 2x⁴ − 7x³ − 2x² + 13x + 6 as a product of four linear factors.

f(x) = 2x^{4} - 7x^{3} - 2x^{2} + 13x + 6 = \{[(2x - 7)x - 2]x + 13\}x + 6

x = -1 \;\Rightarrow\; \{[(2\cdot(-1) - 7)\cdot(-1) - 2]\cdot(-1) + 13\}\cdot(-1) + 6 = 0

x = 2 \;\Rightarrow\; \{[(2\cdot 2 - 7)\cdot 2 - 2]\cdot 2 + 13\}\cdot 2 + 6 = 0

2x^{4} - 7x^{3} - 2x^{2} + 13x + 6 = (x + 1)(x - 2)(ax^{2} + bx + c)

(x + 1)(x − 2) = x² − x − 2

(2x^{4} - 7x^{3} - 2x^{2} + 13x + 6) : (x^{2} - x - 2) = 2x^{2} - 5x - 3

$2x^{4}$	$- 2x^{3}$	$- 4x^{2}$
0	$- 5x^{3}$	$+ 2x^{2}$	$+ 13x$
	$- 5x^{3}$	$+ 5x^{2}$	$+ 10x$
	0	$- 3x^{2}$	$+ 3x$	$+ 6$
		$- 3x^{2}$	$+ 3x$	$+ 6$
				$0$

(x^{2} - x - 2)(2x^{2} - 5x - 3) = 2x^{4} - 7x^{3} - 2x^{2} + 13x + 6

x = \frac{5 \pm \sqrt{(-5)^{2} - 4\cdot 2\cdot(-3)}}{2\cdot 2} = \frac{5 \pm \sqrt{25 + 24}}{4} = \frac{5 \pm 7}{4}

x_{1} = -\tfrac{1}{2}

x_{2} = 3

2x^{2} - 5x - 3 = 2\left(x + \tfrac{1}{2}\right)(x - 3) = (2x + 1)(x - 3)

2x^{4} - 7x^{3} - 2x^{2} + 13x + 6 = (x + 1)(x - 2)(2x + 1)(x - 3)

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Frequently asked questions

How do you make a variable that sits under two roots the subject?

Remove the roots one at a time, squaring both sides at each step. From q = √((5p² − 1)² − 2) you get q² + 2 = (5p² − 1)², then √(q² + 2) = 5p² − 1, and finally p = √((√(q² + 2) + 1)/5).

How many significant figures should the answer keep?

As many as the exercise asks for. Three significant figures turn 1.746424... into 1.75 and 0.387298... into 0.387 - what counts is the number of digits, not decimal places.

How do you quickly check whether x − 2 is a factor?

Evaluate f(2) with the nested form. For 2x³ + 2x² − 17x + 10 we get f(2) = 0, so x − 2 is a factor and the division can be carried out.

What if the quadratic left after the division has no nice factors?

Leave it as a quadratic. The factorisation 2x³ + 2x² − 17x + 10 = (x − 2)(2x² + 6x − 5) is complete over the rationals - the roots of that quadratic are irrational.

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