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Let $f(x) = x^3 + 4x^2 + x - 6.$ (a) Use the factor theorem to show that $(x + 2)$ is a factor of $f(x)$ - Edexcel - A-Level Maths Pure - Question 10 - 2007 - Paper 2

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Let-$f(x)-=-x^3-+-4x^2-+-x---6.$--(a)-Use-the-factor-theorem-to-show-that-$(x-+-2)$-is-a-factor-of-$f(x)$-Edexcel-A-Level Maths Pure-Question 10-2007-Paper 2.png

Let $f(x) = x^3 + 4x^2 + x - 6.$ (a) Use the factor theorem to show that $(x + 2)$ is a factor of $f(x)$. (b) Factorise $f(x)$ completely. (c) Write down all the ... show full transcript

Worked Solution & Example Answer:Let $f(x) = x^3 + 4x^2 + x - 6.$ (a) Use the factor theorem to show that $(x + 2)$ is a factor of $f(x)$ - Edexcel - A-Level Maths Pure - Question 10 - 2007 - Paper 2

Step 1

Use the factor theorem to show that $(x + 2)$ is a factor of $f(x)$

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Answer

To use the factor theorem, we evaluate f(2)f(-2):

f(2)=(2)3+4(2)2+(2)6f(-2) = (-2)^3 + 4(-2)^2 + (-2) - 6

Calculating each term:

  • (2)3=8(-2)^3 = -8
  • 4(2)2=4×4=164(-2)^2 = 4 \times 4 = 16
  • (2)=2(-2) = -2
  • 6=6-6 = -6

Now substituting these values back into the function:

f(2)=8+1626=0f(-2) = -8 + 16 - 2 - 6 = 0

Since f(2)=0f(-2) = 0, it follows that (x+2)(x + 2) is a factor of f(x)f(x).

Step 2

Factorise $f(x)$ completely

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Answer

From part (a), we know that (x+2)(x + 2) is a factor. We can factor f(x)f(x) as:

f(x)=(x+2)(x2+2x3)f(x) = (x + 2)(x^2 + 2x - 3)

Next, we factor the quadratic x2+2x3x^2 + 2x - 3. To do this, we look for two numbers that multiply to 3-3 and add up to 22. These numbers are 33 and 1-1. Therefore:

x2+2x3=(x+3)(x1)x^2 + 2x - 3 = (x + 3)(x - 1)

Putting it all together, we have:

f(x)=(x+2)(x+3)(x1)f(x) = (x + 2)(x + 3)(x - 1)

Step 3

Write down all the solutions to the equation $x^3 + 4x^2 + x - 6 = 0$

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Answer

Setting the factored form equal to zero:

(x+2)(x+3)(x1)=0(x + 2)(x + 3)(x - 1) = 0

This gives us the solutions:

  • x+2=0x=2x + 2 = 0 \Rightarrow x = -2
  • x+3=0x=3x + 3 = 0 \Rightarrow x = -3
  • x1=0x=1x - 1 = 0 \Rightarrow x = 1

Thus, the solutions are:

x=3,2,1x = -3, -2, 1

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