Solved on Mar 22, 2024

(a) Solve $2(x+2)-5=9$ . (b) Write as single fraction $\frac{2x+1}{3}+\frac{3x-2}{6}$ . (c) Rearrange $T=2\pi\sqrt{\frac{L}{8}}$ to find $L$ . (d) (I) Show $f(x)=x^3-13x+12$ can be written as $(x-1)(x^2+x-12)$ . (II) Completely factorise $f(x)$ .

STEP 1

Assumptions for part (a)
1. We are given the equation $2(x+2)-5=9$ .
2. We need to solve for $x$ .

STEP 2

First, we will distribute the 2 into the parentheses.
$2(x+2) - 5 = 9$
$2x + 4 - 5 = 9$

STEP 3

Combine like terms on the left side of the equation.
$2x - 1 = 9$

STEP 4

Add 1 to both sides of the equation to isolate the term with $x$ .
$2x - 1 + 1 = 9 + 1$
$2x = 10$

STEP 5

Divide both sides by 2 to solve for $x$ .
$\frac{2x}{2} = \frac{10}{2}$
$x = 5$
So, the solution to the equation $2(x+2)-5=9$ is $x = 5$ .

STEP 6

Assumptions for part (b)
1. We are given two fractions $\frac{2x+1}{3}$ and $\frac{3x-2}{6}$ .
2. We need to combine them into a single fraction.

STEP 7

Find a common denominator for the two fractions. The least common multiple of 3 and 6 is 6.

STEP 8

Adjust the first fraction so that it has the common denominator of 6.
$\frac{2x+1}{3} = \frac{2x+1}{3} \cdot \frac{2}{2} = \frac{4x+2}{6}$

STEP 9

Now that both fractions have the same denominator, we can add them together.
$\frac{4x+2}{6} + \frac{3x-2}{6}$

STEP 10

Combine the numerators over the common denominator.
$\frac{(4x+2) + (3x-2)}{6}$

STEP 11

Combine like terms in the numerator.
$\frac{4x + 3x + 2 - 2}{6}$
$\frac{7x}{6}$
So, the single fraction is $\frac{7x}{6}$ .

STEP 12

Assumptions for part (c)
1. We are given the formula for the period of a pendulum $T=2\pi\sqrt{\frac{L}{8}}$ .
2. We need to rearrange the formula to solve for $L$ .

STEP 13

First, we will square both sides of the equation to eliminate the square root.
$T^2 = (2\pi\sqrt{\frac{L}{8}})^2$
$T^2 = 4\pi^2\frac{L}{8}$

STEP 14

Multiply both sides by $\frac{8}{4\pi^2}$ to isolate $L$ .
$\frac{8}{4\pi^2}T^2 = L$

STEP 15

Simplify the equation.
$L = \frac{2T^2}{\pi^2}$
So, the rearranged formula with $L$ as the subject is $L = \frac{2T^2}{\pi^2}$ .

STEP 16

Assumptions for part (d)(I)
1. We are given the cubic function $f(x) = x^3 - 13x + 12$ .
2. We need to show that $f(x)$ can be written as $(x-1)(x^2+x-12)$ .

STEP 17

First, we will attempt to factor by grouping. Look for a common factor in the terms of $f(x)$ .

STEP 18

Notice that $x=1$ is a root of the equation $f(x) = 0$ because $f(1) = 1^3 - 13(1) + 12 = 0$ .

STEP 19

Since $x=1$ is a root, $(x-1)$ is a factor of $f(x)$ .

STEP 20

Perform polynomial division or use synthetic division to divide $f(x)$ by $(x-1)$ .

STEP 21

Using synthetic division with the root $x=1$ :
$\begin{array}{c|ccc} 1 & 1 & 0 & -13 & 12 \\ & & 1 & 1 & -12 \\ \hline & 1 & 1 & -12 & 0 \\ \end{array}$

STEP 22

The result of the division gives us the quadratic $x^2 + x - 12$ .

STEP 23

Therefore, $f(x)$ can be written as $(x-1)(x^2+x-12)$ .

STEP 24

Assumptions for part (d)(II)
1. We have shown that $f(x) = (x-1)(x^2+x-12)$ .
2. We need to factorise $f(x)$ completely.

STEP 25

Now, we will factor the quadratic $x^2 + x - 12$ .

STEP 26

Look for two numbers that multiply to -12 and add to 1. These numbers are 4 and -3.

STEP 27

Factor the quadratic as $(x+4)(x-3)$ .

STEP 28

Thus, the complete factorisation of $f(x)$ is $(x-1)(x+4)(x-3)$ .
The solutions to the parts of the problem are as follows:
(a) $x = 5$
(b) $\frac{7x}{6}$
(c) $L = \frac{2T^2}{\pi^2}$
(d)(I) $f(x) = (x-1)(x^2+x-12)$
(d)(II) $f(x) = (x-1)(x+4)(x-3)$

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STEP 1

Assumptions for part (a)1. We are given the equation 2(x+2)−5=92(x+2)-5=92(x+2)−5=9.2. We need to solve for xxx.

STEP 2

First, we will distribute the 2 into the parentheses.2(x+2)−5=92(x+2) - 5 = 92(x+2)−5=92x+4−5=92x + 4 - 5 = 92x+4−5=9

STEP 3

Combine like terms on the left side of the equation.2x−1=92x - 1 = 92x−1=9

STEP 4

Add 1 to both sides of the equation to isolate the term with xxx.2x−1+1=9+12x - 1 + 1 = 9 + 12x−1+1=9+12x=102x = 102x=10

STEP 5

Divide both sides by 2 to solve for xxx.2x2=102\frac{2x}{2} = \frac{10}{2}22x​=210​x=5x = 5x=5So, the solution to the equation 2(x+2)−5=92(x+2)-5=92(x+2)−5=9 is x=5x = 5x=5.

STEP 6

Assumptions for part (b)1. We are given two fractions 2x+13\frac{2x+1}{3}32x+1​ and 3x−26\frac{3x-2}{6}63x−2​.2. We need to combine them into a single fraction.

STEP 7

Find a common denominator for the two fractions. The least common multiple of 3 and 6 is 6.

STEP 8

Adjust the first fraction so that it has the common denominator of 6.2x+13=2x+13⋅22=4x+26\frac{2x+1}{3} = \frac{2x+1}{3} \cdot \frac{2}{2} = \frac{4x+2}{6}32x+1​=32x+1​⋅22​=64x+2​

STEP 9

Now that both fractions have the same denominator, we can add them together.4x+26+3x−26\frac{4x+2}{6} + \frac{3x-2}{6}64x+2​+63x−2​

STEP 10

Combine the numerators over the common denominator.(4x+2)+(3x−2)6\frac{(4x+2) + (3x-2)}{6}6(4x+2)+(3x−2)​

STEP 11

Combine like terms in the numerator.4x+3x+2−26\frac{4x + 3x + 2 - 2}{6}64x+3x+2−2​7x6\frac{7x}{6}67x​So, the single fraction is 7x6\frac{7x}{6}67x​.

STEP 12

Assumptions for part (c)1. We are given the formula for the period of a pendulum T=2πL8T=2\pi\sqrt{\frac{L}{8}}T=2π8L​​.2. We need to rearrange the formula to solve for LLL.

STEP 13

First, we will square both sides of the equation to eliminate the square root.T2=(2πL8)2T^2 = (2\pi\sqrt{\frac{L}{8}})^2T2=(2π8L​​)2T2=4π2L8T^2 = 4\pi^2\frac{L}{8}T2=4π28L​

STEP 14

Multiply both sides by 84π2\frac{8}{4\pi^2}4π28​ to isolate LLL.84π2T2=L\frac{8}{4\pi^2}T^2 = L4π28​T2=L

STEP 15

Simplify the equation.L=2T2π2L = \frac{2T^2}{\pi^2}L=π22T2​So, the rearranged formula with LLL as the subject is L=2T2π2L = \frac{2T^2}{\pi^2}L=π22T2​.

STEP 16

Assumptions for part (d)(I)1. We are given the cubic function f(x)=x3−13x+12f(x) = x^3 - 13x + 12f(x)=x3−13x+12.2. We need to show that f(x)f(x)f(x) can be written as (x−1)(x2+x−12)(x-1)(x^2+x-12)(x−1)(x2+x−12).

STEP 17

First, we will attempt to factor by grouping. Look for a common factor in the terms of f(x)f(x)f(x).

STEP 18

Notice that x=1x=1x=1 is a root of the equation f(x)=0f(x) = 0f(x)=0 because f(1)=13−13(1)+12=0f(1) = 1^3 - 13(1) + 12 = 0f(1)=13−13(1)+12=0.

STEP 19

Since x=1x=1x=1 is a root, (x−1)(x-1)(x−1) is a factor of f(x)f(x)f(x).

STEP 20

Perform polynomial division or use synthetic division to divide f(x)f(x)f(x) by (x−1)(x-1)(x−1).

STEP 21

Using synthetic division with the root x=1x=1x=1:110−131211−1211−120\begin{array}{c|ccc} 1 & 1 & 0 & -13 & 12 \\ & & 1 & 1 & -12 \\ \hline & 1 & 1 & -12 & 0 \\ \end{array}1​11​011​−131−12​12−120​​

STEP 22

The result of the division gives us the quadratic x2+x−12x^2 + x - 12x2+x−12.

STEP 23

Therefore, f(x)f(x)f(x) can be written as (x−1)(x2+x−12)(x-1)(x^2+x-12)(x−1)(x2+x−12).

STEP 24

Assumptions for part (d)(II)1. We have shown that f(x)=(x−1)(x2+x−12)f(x) = (x-1)(x^2+x-12)f(x)=(x−1)(x2+x−12).2. We need to factorise f(x)f(x)f(x) completely.

STEP 25

Now, we will factor the quadratic x2+x−12x^2 + x - 12x2+x−12.

STEP 26

Look for two numbers that multiply to -12 and add to 1. These numbers are 4 and -3.

STEP 27

Factor the quadratic as (x+4)(x−3)(x+4)(x-3)(x+4)(x−3).

STEP 28

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Assumptions for part (a)
1. We are given the equation $2(x+2)-5=9$ .
2. We need to solve for $x$ .

First, we will distribute the 2 into the parentheses.
$2(x+2) - 5 = 9$
$2x + 4 - 5 = 9$

Combine like terms on the left side of the equation.
$2x - 1 = 9$

Add 1 to both sides of the equation to isolate the term with $x$ .
$2x - 1 + 1 = 9 + 1$
$2x = 10$

Divide both sides by 2 to solve for $x$ .
$\frac{2x}{2} = \frac{10}{2}$
$x = 5$
So, the solution to the equation $2(x+2)-5=9$ is $x = 5$ .

Assumptions for part (b)
1. We are given two fractions $\frac{2x+1}{3}$ and $\frac{3x-2}{6}$ .
2. We need to combine them into a single fraction.

Adjust the first fraction so that it has the common denominator of 6.
$\frac{2x+1}{3} = \frac{2x+1}{3} \cdot \frac{2}{2} = \frac{4x+2}{6}$

Now that both fractions have the same denominator, we can add them together.
$\frac{4x+2}{6} + \frac{3x-2}{6}$

Combine the numerators over the common denominator.
$\frac{(4x+2) + (3x-2)}{6}$

Combine like terms in the numerator.
$\frac{4x + 3x + 2 - 2}{6}$
$\frac{7x}{6}$
So, the single fraction is $\frac{7x}{6}$ .

Assumptions for part (c)
1. We are given the formula for the period of a pendulum $T=2\pi\sqrt{\frac{L}{8}}$ .
2. We need to rearrange the formula to solve for $L$ .

First, we will square both sides of the equation to eliminate the square root.
$T^2 = (2\pi\sqrt{\frac{L}{8}})^2$
$T^2 = 4\pi^2\frac{L}{8}$

Multiply both sides by $\frac{8}{4\pi^2}$ to isolate $L$ .
$\frac{8}{4\pi^2}T^2 = L$

Simplify the equation.
$L = \frac{2T^2}{\pi^2}$
So, the rearranged formula with $L$ as the subject is $L = \frac{2T^2}{\pi^2}$ .

Assumptions for part (d)(I)
1. We are given the cubic function $f(x) = x^3 - 13x + 12$ .
2. We need to show that $f(x)$ can be written as $(x-1)(x^2+x-12)$ .

First, we will attempt to factor by grouping. Look for a common factor in the terms of $f(x)$ .

Notice that $x=1$ is a root of the equation $f(x) = 0$ because $f(1) = 1^3 - 13(1) + 12 = 0$ .

Since $x=1$ is a root, $(x-1)$ is a factor of $f(x)$ .

Perform polynomial division or use synthetic division to divide $f(x)$ by $(x-1)$ .

Using synthetic division with the root $x=1$ :
$\begin{array}{c|ccc} 1 & 1 & 0 & -13 & 12 \\ & & 1 & 1 & -12 \\ \hline & 1 & 1 & -12 & 0 \\ \end{array}$

The result of the division gives us the quadratic $x^2 + x - 12$ .

Therefore, $f(x)$ can be written as $(x-1)(x^2+x-12)$ .

Assumptions for part (d)(II)
1. We have shown that $f(x) = (x-1)(x^2+x-12)$ .
2. We need to factorise $f(x)$ completely.

Now, we will factor the quadratic $x^2 + x - 12$ .

Factor the quadratic as $(x+4)(x-3)$ .