Let $P(x) = x^3 - ax^2 + x + b$ be a polynomial, where $a$ is a real number - HSC - SSCE Mathematics Extension 1 - Question 2 - 2011 - Paper 1
Question 2
Let $P(x) = x^3 - ax^2 + x + b$ be a polynomial, where $a$ is a real number.
When $P(x)$ is divided by $x - 3$ the remainder is 12.
Find the remainder when $P(x)$ ... show full transcript
Worked Solution & Example Answer:Let $P(x) = x^3 - ax^2 + x + b$ be a polynomial, where $a$ is a real number - HSC - SSCE Mathematics Extension 1 - Question 2 - 2011 - Paper 1
Step 1
When $P(x)$ is divided by $x + 1$
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Answer
To find the remainder when P(x) is divided by x+1, we can use the Remainder Theorem, which states that the remainder of the division of a polynomial P(x) by x−c is given by P(c).
Substitute x=−1 into the polynomial.
We need to first find P(−1):
P(−1)=(−1)3−a(−1)2+(−1)+b
This simplifies to:
P(−1)=−1−a−1+b=b−a−2
We know that P(3)=12 from the division by x−3:
P(3)=33−a(32)+3+b
This simplifies to:
27−9a+3+b=12
So, we solve for b,
ightarrow b = 12 - 30 + 9a = 9a - 18$$
Substitute b back into P(−1):
P(−1)=(9a−18)−a−2=8a−20
Thus, the remainder when P(x) is divided by x+1 is 8a−20.
Step 2
Newton's method to obtain another approximation
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Answer
We start with the function f(x)=extcos(2x)−x, with a known zero near x = rac{1}{2}.
Calculate the derivative:
f′(x)=−2extsin(2x)−1
Use Newton's method formula:
xn+1=xn−f′(xn)f(xn)
Substitute x0=0.5 into f(x):
f(0.5)=extcos(1)−0.5
Compute extcos(1) using a calculator, approximately 0.5403:
f(0.5)=0.5403−0.5≈0.0403
Now find f′(0.5):
f′(0.5)=−2extsin(1)−1≈−2(0.8415)−1≈−2.683
Plug values into Newton's method:
x1=0.5−−2.6830.0403≈0.515
Finally, rounded to two decimal places, we find: x1≈0.52.
Step 3
Find an expression for the coefficient of $x^2$
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Answer
To find the coefficient of x2 in the expansion of (3x−4−x8)8, we can use the binomial theorem:
(a+b)n=∑k=0n(kn)an−kbk
Setting a=3x−4 and b=−x8, we need to find the term involving x2: