Help with some methods questions 3/4!

[hello_kitty]

Hi Guys,
Ive been doing chapter 1 of methods 3/4 and have some questions! any help would be great
Im a bi confused on the Remainder Theorem..could someone please explain it to me?

Also, I have these questions

1)   The polynomial x³ + px² + qx + 6 is exactly divisble by (x+2) and (x-3). Find the values of p and q

Answer: p = -2, q = -5


2a) Factorise        2x³ - 18x² + 54x - 54

I got up to   2 (x³ - 9x² + 47x - 47)

Answer: 2 (x-3)³



2b) 3x³ + 9x² - 18x - 24

Answer: 3 (x+4)(x-2)(x+1)


3) If x³ + 4x² - 3x + 2 = x³ + (mx+n)² + 5 or all values of x, find the values of m and n

Answer: m = 2 , n = - 3  OR  m = -2, n = 3

4a) If x² = a (x+2)² + b (x +2) + c for all values of x, Find the values of a,b and c
Answer:  a = 1, b = -4, c = 4


b)   If 3x² - 12x + 11 = A (x-h)² + k, Find the values of A, h, and k

Answer: A = 3, h = 2, k= -1


Thanks for any help

[Special At Specialist]

1) x^3 + px^2 + qx + 6
Divisible by (x + 2) and (x - 3), meaning the remainder is 0 when you divide by either of these.
Let's start by dividing by (x + 2) (I like to use synthetic division but long division is also fine):
-2 [1     p     q         6]
    [0   -2   4-2p   4p-8+2q]
     1  p-2  4-2p+q    2q+4p-2
So the remainder is (2q + 4p - 2)
But we know that the remainder is zero, so:
2q + 4p - 2 = 0
We can do the same for the other one and let that equal 0, and then we just solve them like simultaneous equations.

2a) 2x^3 - 18x^2 + 54x - 54
They are all even numbers, so factor out a 2:
= 2(x^3 - 9x^2 + 27x - 27)
27 is divisible by both 3 and 9, so that should be a hint to use the formula:
(a - b)^3 = a^3 - 3a^2b + 3ab^2 - b^3
2(x^3 - 9x^2 + 27x - 27)
= 2(x^3 - 3(3)(x)^2 + 3(3)^2(x) - (3)^3)
= 2(x - 3)^3

2b) 3x^3 + 9x^2 - 18x - 24
They are all divisible by 3, so factor out a 3:
= 3(x^3 + 3x^2 - 6x -
Now test numbers using the remainder and factor theorem.
Let P(x) = x^3 + 3x^2 - 6x - 8
Ignore the 3 outside the brackets:
P(0) = 0 + 0 - 0 - 8 ≠ 0
P(1) = 1 + 3 - 6 - 8 ≠ 0
P(2) = 8 + 12 - 12 - 8 = 0
So (x - 2) is a factor.
Now divide the polynomial P(x) by (x - 2):
2 [1     3     -6     -8]
   [0     2     10     8]
    1     5      4      0
P(x) / (x - 2) = x^2 + 5x + 4
So the entire thing is:
2(x - 2)(x^2 + 5x + 4)
We can factorise x^2 + 5x + 4 into (x + 1)(x + 4)
= 2(x - 2)(x + 1)(x + 4)

3. x^3 + 4x^2 - 3x + 2 = x^3 + (mx + n)^2 + 5
This is just a matter of expanding and equating coefficients:
RHS = x^3 + (mx)^2 + 2(mx)(n) + n^2 + 5
RHS = x^3 + (m^2)x^2 + (2mn)x + (n^2 + 5)
So let the x^3 terms equal the x^3 terms, the x^2 terms equal the x^2 terms etc:
1 = 1
m^2 = 4
2mn = -3
n^2 + 5 = 2
n^2 = -3
No real solution for n.
Something must be wrong with this equation... perhaps I read it wrong?

4a) x^2 = a(x + 2)^2 + b(x + 2) + c
RHS = a(x^2 + 4x + 4) + bx + 2b + c
RHS = ax^2 + 4ax + 4a + bx + 2b + c
RHS = (a)x^2 + (4a + b)x + (4a + 2b + c)
Equating coefficients:
a = 1
4a + b = 0
4a + 2b + c = 0
Solve these equations:
4 + b = 0 so b = -4
4 - 8 + c = 0 so c = 4
a = 1, b = -4 and c = 4

b) 3x^2 - 12x + 11 = A(x - h)^2 + k
RHS = A(x^2 - 2xh + h^2) + k
RHS = Ax^2 - 2Ahx + Ah^2 + k
RHS = (A)x^2 + (-2AH)x + (Ah^2 + k)
Equate coefficients:
A = 3
-2AH = -12
Ah^2 + k = 11
Solve them:
-6h = -12 so h = 2
3*4 + k = 11 so k = -1
A = 3, h = 2 and k = -1

[hello_kitty]

You are amazing! thank you so much!