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4 (a) Use the factor theorem to prove that .x - 6 - 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 - 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).
Answer
To prove that .x - 6. is a factor of .p(x), we will substitute .x = 6. into .p(x).
Calculate:
First, compute each term:
4(6)^3 &= 4 imes 216 = 864, \ -15(6)^2 &= -15 imes 36 = -540, \ -48(6) &= -288, \ -36 &= -36. \end{align*}$$ Now summing these values: $$p(6) = 864 - 540 - 288 - 36 = 0.$$ Since .p(6) = 0., by the factor theorem, we conclude that .x - 6. is a factor of .p(x).Step 2
Prove that the graph of .y = p(x). intersects the .x-axis. at exactly one point.
Answer
To prove that .y = p(x). intersects the .x-axis at exactly one point, we need to analyze the roots of .p(x).
First, we factor .p(x):
Knowing that .x - 6. is a factor, we can divide .p(x). by .x - 6. to find the other factors or confirm there are no other real roots.
Carrying out synthetic division or polynomial long division, we get:
We then find that the quadratic .Ax^2 + Bx + C. has a discriminant given by:
If we find that the discriminant is negative, then .p(x). has exactly one real root corresponding to .x - 6., verifying our statement: if .D < 0., then .p(x). intersects the x-axis only once.
Step 3