A-Level AQA Maths Pure Graphs of Functions: 4 (a) Use the factor theorem to

4 (a) Use the factor theorem to prove that $x - 6$ is a factor of $p(x)$ - AQA - A-Level Maths Pure - Question 4 - 2020 - Paper 3

Question 4

4 (a) Use the factor theorem to prove that $x - 6$ is a factor of $p(x)$. 4 (b) (i) Prove that the graph of $y = p(x)$ intersects the x-axis at exactly one point. ... show full transcript

Worked Solution & Example Answer:4 (a) Use the factor theorem to prove that $x - 6$ is a factor of $p(x)$ - AQA - A-Level Maths Pure - Question 4 - 2020 - Paper 3

Step 1

Use the factor theorem to prove that $x - 6$ is a factor of $p(x)$.

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Answer

To use the factor theorem, we substitute $x = 6$ into the polynomial $p(x) = 4x^3 - 15x^2 - 48x - 36$ . First, we calculate:

$p(6) = 4(6)^3 - 15(6)^2 - 48(6) - 36$

Calculating each term:

$4(6)^3 = 864$
$-15(6)^2 = -540$
$-48(6) = -288$
$-36 = -36$

Adding these together:

$p(6) = 864 - 540 - 288 - 36 = 0$

Since $p(6) = 0$ , this implies that $x - 6$ is indeed a factor of $p(x)$ .

Step 2

Prove that the graph of $y = p(x)$ intersects the x-axis at exactly one point.

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Answer

To determine where the graph of $y = p(x)$ intersects the x-axis, we set $p(x) = 0$ . Since we established in part (a) that $x - 6$ is a factor, we can factor the polynomial:

$p(x) = (x - 6)(4x^2 + 9x + 6)$

Next, we need to analyze the quadratic factor $4x^2 + 9x + 6$ by calculating its discriminant:

$D = b^2 - 4ac = (9)^2 - 4(4)(6) = 81 - 96 = -15$

Since the discriminant is less than zero ( $D < 0$ ), this quadratic has no real roots. Therefore, the only intersection point of the graph with the x-axis occurs at $x = 6$ . Hence, the graph intersects the x-axis at exactly one point.

Step 3

State the coordinates of this point of intersection.

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Answer

The coordinates of the point of intersection are $(6, 0)$ .

4 (a) Use the factor theorem to prove that $x - 6$ is a factor of $p(x)$ - AQA - A-Level Maths Pure - Question 4 - 2020 - Paper 3

Worked Solution & Example Answer:4 (a) Use the factor theorem to prove that $x - 6$ is a factor of $p(x)$ - AQA - A-Level Maths Pure - Question 4 - 2020 - Paper 3

Use the factor theorem to prove that $x - 6$ is a factor of $p(x)$.

Prove that the graph of $y = p(x)$ intersects the x-axis at exactly one point.

State the coordinates of this point of intersection.

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