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Given that P(x) = 125x^3 + 150x^2 + 55x + 6 use the factor theorem to prove that (5x + 1) is a factor of P(x) - AQA - A-Level Maths Pure - Question 13 - 2021 - Paper 1

Question 13

Given that P(x) = 125x^3 + 150x^2 + 55x + 6 use the factor theorem to prove that (5x + 1) is a factor of P(x). Factorise P(x) completely. Hence, prove that 250n^3 ... show full transcript

Worked Solution & Example Answer:Given that P(x) = 125x^3 + 150x^2 + 55x + 6 use the factor theorem to prove that (5x + 1) is a factor of P(x) - AQA - A-Level Maths Pure - Question 13 - 2021 - Paper 1

Step 1

use the factor theorem to prove that (5x + 1) is a factor of P(x)

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Answer

To apply the factor theorem, we first substitute the root of the factor, which is found by solving the equation: $5x + 1 = 0$ . This gives us the root: $x = -\frac{1}{5}$ . Next, we substitute this value into the polynomial:

$P(-\frac{1}{5}) = 125\left(-\frac{1}{5}\right)^3 + 150\left(-\frac{1}{5}\right)^2 + 55\left(-\frac{1}{5}\right) + 6$

Calculating each term:

First term: $125\left(-\frac{1}{5}\right)^3 = 125\left(-\frac{1}{125}\right) = -1$
Second term: $150\left(-\frac{1}{5}\right)^2 = 150\left(\frac{1}{25}\right) = 6$
Third term: $55\left(-\frac{1}{5}\right) = -11$
Fourth term: $6 = 6$

Now, summing these values: $P(-\frac{1}{5}) = -1 + 6 - 11 + 6 = 0$

Since $P(-\frac{1}{5}) = 0$ , it follows that $(5x + 1)$ is indeed a factor of $P(x)$ .

Step 2

Factorise P(x) completely

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Answer

Given $P(x) = 125x^3 + 150x^2 + 55x + 6$ , we know that $(5x + 1)$ is a factor. We can perform polynomial long division to find the other factor:

Dividing $P(x)$ by $(5x + 1)$ , we find: $P(x) = (5x + 1)(25x^2 + 5x + 6)$

Next, we factor the quadratic $25x^2 + 5x + 6$ . To factor this, we can use the quadratic formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\ ext{ where } a = 25, b = 5, c = 6$ This leads to: $x = \frac{-5 \pm \sqrt{5^2 - 4 \cdot 25 \cdot 6}}{2 \cdot 25} = \frac{-5 \pm \sqrt{25 - 600}}{50} = \frac{-5 \pm \sqrt{-575}}{50}$

The roots are complex, indicating we cannot factor further over the reals. Therefore, we conclude that the complete factorization is: $P(x) = (5x + 1)(25x^2 + 5x + 6)$

Step 3

prove that 250n^3 + 300n^2 + 110n + 12 is a multiple of 12 when n is a positive whole number

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Answer

First, we will factor $250n^3 + 300n^2 + 110n + 12$ :

$250n^3 + 300n^2 + 110n + 12 = 250n^3 + 300n^2 + 110n + 12 = 2(5n^3 + 6n^2 + 5n + 6)$

Next, we check if $5n^3 + 6n^2 + 5n + 6$ can be structured to follow the divisibility rule of 12. Noting that:

$5n, 6n^2, 110n,$ and 12 are multiples contributing individual factors:

$5n + 1, 5n + 2, 5n + 3$

This shows that they are consecutive integer summations. The factors formed produce results that are multiples of 3 and contain numbers that contribute even divisibility by 2.

Therefore, we can conclude: $ext{Since } 250n^3 + 300n^2 + 110n + 12 ext{ is formed by multiples of 6 (each term contributing), it is indeed a multiple of 12.}$

Given that P(x) = 125x^3 + 150x^2 + 55x + 6 use the factor theorem to prove that (5x + 1) is a factor of P(x) - AQA - A-Level Maths Pure - Question 13 - 2021 - Paper 1

Worked Solution & Example Answer:Given that P(x) = 125x^3 + 150x^2 + 55x + 6 use the factor theorem to prove that (5x + 1) is a factor of P(x) - AQA - A-Level Maths Pure - Question 13 - 2021 - Paper 1

use the factor theorem to prove that (5x + 1) is a factor of P(x)

Factorise P(x) completely

prove that 250n^3 + 300n^2 + 110n + 12 is a multiple of 12 when n is a positive whole number

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