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(a) (i) Show that \((x+2)\) is a factor of \(f(x) = x^3 - 2x^2 - 20x - 24\) - Scottish Highers Maths - Question 13 - 2022
Question 13
(a) (i) Show that \((x+2)\) is a factor of \(f(x) = x^3 - 2x^2 - 20x - 24\). (ii) Hence, or otherwise, solve \(f(x) = 0\). The diagram shows the graph of \(y = f(x)... show full transcript
Worked Solution & Example Answer:(a) (i) Show that \((x+2)\) is a factor of \(f(x) = x^3 - 2x^2 - 20x - 24\) - Scottish Highers Maths - Question 13 - 2022
Step 1
Show that \((x+2)\) is a factor of \(f(x) = x^3 - 2x^2 - 20x - 24\)
Answer
To show that ((x+2)) is a factor, we will use synthetic division. We substitute (-2) into the polynomial:
-
Set up synthetic division: -2 | 1 -2 -20 -24
|
---------------- -
Perform synthetic division: -2 | 1 -2 -20 -24
| -2 8 24
---------------- (0 ; )
The remainder is (0), indicating that ((x + 2)) is a factor of (f(x)).
Step 2
Hence, or otherwise, solve \(f(x) = 0\)
Answer
We can factor (f(x)) as follows using the known factor ((x+2)):
[ f(x) = (x + 2)(x^2 - 4x - 12) ]
Next, we apply the quadratic formula to (x^2 - 4x - 12 = 0):
[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{4 \pm \sqrt{(-4)^2 - 4 \cdot 1 \cdot (-12)}}{2 \cdot 1} ]
[ x = \frac{4 \pm \sqrt{16 + 48}}{2} = \frac{4 \pm \sqrt{64}}{2} = \frac{4 \pm 8}{2} ]
This gives us the solutions (x = 6) and (x = -2).
Thus, (f(x) = 0) has the solutions: (-2, 6).
Step 3
State the value of \(k\)
Answer
Since the graph of (y = f(x - k)) has a stationary point at ((1, 0)), we deduce that the function has been shifted. The stationary point occurs when:
[ f'(1) = 0 ]
To find (k), note that the stationary point indicates that (x - k = 1): therefore, (k = 1 - 0 = 1).
Hence, the value of (k) is (3).