Photo AI

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 Mechanics - 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 t... 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 Mechanics - Question 13 - 2021 - Paper 1

Step 1

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

96%

114 rated

Answer

To use the factor theorem, we substitute $x = -\frac{1}{5}$ into $P(x)$ .

Calculating:

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

= 125\left(-\frac{1}{125}\right) + 150\left(\frac{1}{25}\right) - 11 - 6

= -1 + 6 - 11 - 6 = 0

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

Step 2

Factorise P(x) completely

99%

104 rated

Answer

We know that one factor is $(5x + 1)$ . We can perform polynomial long division of $P(x)$ by $(5x + 1)$ :

Divide the leading term: $125x^3 \div 5x = 25x^2$ .
Multiply: $(5x + 1)(25x^2) = 125x^3 + 25x^2$ .
Subtract: $P(x) - (125x^3 + 25x^2)$ results in $125x^2 + 55x - 6$ .
Repeat with $125x^2 \div 5x = 25x$ .
Multiply: $(5x + 1)(25x) = 125x^2 + 25x$ .
Subtract: results in $30x - 6$ .
Finally, $30x \div 5x = 6$ and multiply to find the next factor.

Hence, from the factors found we get:

P(x) = (5x + 1)(5x + 2)(5x + 3)

Step 3

prove that 250n^3 + 300n^2 + 110n + 12 is a multiple of 12

96%

101 rated

Answer

From the established factors, substituting yields:

250n^3 + 300n^2 + 110n + 12 = 2(5n)(5n + 1)(5n + 2)

The terms $(5n)$ , $(5n + 1)$ , and $(5n + 2)$ are three consecutive integers where at least one of them must be a multiple of 3, and at least one is even. Therefore, the product contains a multiple of 2 and 3, ensuring that:

12 | \left(250n^3 + 300n^2 + 110n + 12\right)

Thus, the expression evaluates as a multiple of 12 when $n$ is a positive whole number.

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 Mechanics - 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 Mechanics - 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

Join the A-Level students using SimpleStudy...