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 ... 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 use the factor theorem, we substitute x=−51 into P(x):
P(−51)=125(−51)3+150(−51)2+55(−51)+6
Calculating each term, we get:
First term: 125⋅−1251=−1
Second term: 150⋅251=6
Third term: 55⋅−51=−11
Constant term: 6
Thus:
P(−51)=−1+6−11+6=0
Since P(−51)=0, by the factor theorem, (5x+1) is indeed a factor of P(x).
Step 2
Factorise $P(x)$ completely
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Answer
We can start with the factor (5x+1) and divide P(x) by it to find the other factors:
Performing synthetic division:
The resulting polynomial is:
P(x)=(5x+1)(Ax2+Bx+C)
After finding the quadratic factor, we can factorize:
P(x)=(5x+1)(5x+2)(x+3)
Step 3
prove that $250n^3 + 300n^2 + 110n + 12$ is a multiple of 12
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Answer
We start with:
250n3+300n2+110n+12=2(125n3+150n2+55n+6)
Next, we factor out:
=2(5n+1)(5n+2)(5n+3)
The factors (5n+1), (5n+2), and (5n+3) are three consecutive whole numbers. Among any three consecutive integers, at least one is a multiple of 3 and at least one is even, meaning:
This guarantees the product (5n+1)(5n+2)(5n+3) is a multiple of 2 and 3.
Hence:
250n3+300n2+110n+12 is a multiple of 2⋅3=6, and as it is multiplied by 2, it also confirms it is a multiple of 12.
