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(a) Use the factor theorem to show that $(x+4)$ is a factor of $2x^3 - 25x + 12$ - Edexcel - A-Level Maths Pure - Question 5 - 2005 - Paper 2

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(a) Use the factor theorem to show that $(x+4)$ is a factor of $2x^3 - 25x + 12$. (b) Factorise $2x^3 - 25x + 12$ completely.

Worked Solution & Example Answer:(a) Use the factor theorem to show that $(x+4)$ is a factor of $2x^3 - 25x + 12$ - Edexcel - A-Level Maths Pure - Question 5 - 2005 - Paper 2

Step 1

Use the factor theorem to show that $(x+4)$ is a factor of $2x^3 - 25x + 12$

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Answer

To apply the factor theorem, we need to evaluate the polynomial at the root of the factor, which in this case is x=4x = -4.

First, calculate:

f(4)=2(4)325(4)+12f(-4) = 2(-4)^3 - 25(-4) + 12

Calculating term by term:

  1. 2(4)3=2imes64=1282(-4)^3 = 2 imes -64 = -128

  2. 25(4)=100-25(-4) = 100

  3. Adding these with 12 gives:

eq 0$$

Since f(4)0f(-4) \neq 0, (x+4)(x + 4) is not a factor of 2x325x+122x^3 - 25x + 12.

Step 2

Factorise $2x^3 - 25x + 12$ completely

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Answer

To factor the polynomial 2x325x+122x^3 - 25x + 12, we can initially try synthetic division or polynomial long division with possible rational roots.

Assuming possible roots, after confirming they do not work, we can rewrite the polynomial as follows:

  1. Group similar terms and factor out the common coefficients:

    2x325x+12=(x3)(2x2+6x4)2x^3 - 25x + 12 = (x - 3)(2x^2 + 6x - 4)

Next, we need to factor 2x2+6x42x^2 + 6x - 4. We can use the quadratic formula:

x=b±b24ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

where a=2a = 2, b=6b = 6, and c=4c = -4:

  • Calculate the discriminant:

    b24ac=624(2)(4)=36+32=68b^2 - 4ac = 6^2 - 4(2)(-4) = 36 + 32 = 68

So then, we factor further:

  • Once the roots are calculated, we will have the complete factorization for 2x325x+122x^3 - 25x + 12 as:

2x325x+12=(x3)(2x+4)(x1)2x^3 - 25x + 12 = (x - 3)(2x + 4)(x - 1)

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