Maths Help PLease?

Question

Maths Questions Pretty Please..?
Could anyone please help me out, with the full working out and solution to each problem. thanks

QUESTION 2) Use the remainder theorem to find the remainder when 2x^4 - x^2 + 3x - 2 is divided by (2x-1)

QUESTION 3) When x^4 + ax^2 + bx - 2 is divided by (x-1) the remainder is -2, and when it is divided by (x+1) the remainder is -6. Find the values of a and b.

QUESTION 5) The polynomial x^3 +px^2+6 is exactly divisible by (x+2) and (x-3). Find the values or p and q.

QUESTION 7) If x^3 + 4x^2 - 3x +2 = (x+1)(ax^2+bx+c) + d for all the values of x, find the values of a,b,c, and d.

QUESTION 8) If x^3 + 4x^2 - 12x + 14 = x^3 + (mx + n)^2 + 5 for all the values of x, find the values of m and n.

QUESTION 9) If x^2 = a(x+2)^2 + b(x+2) + c for all values of x, find the values of a,b,c.

THANKS HEAPS IN ADVANCE..

ops.. sorry Question 5 i wrote wrong, i left out the q... its;

Question 5) x^3 +px^2 + qx + 6

sorry

Answer 1

3) Let p(x) = x^4 + ax^2 + bx - 2

We're given the remainder, when divided by (x - 1) is -2. This means p(1) = -2 (i.e. if (x - r) is a factor, r is the remainder).

But p(1) = 1^4 + a(1)^2 + b(1) - 2, so
p(1) = 1 + a + b - 2
p(1) = a + b - 1

But p(1) = -2, so

a + b - 1 = -2
a + b = -1

When it is divided by (x + 1), the remainder is -6, which means
p(-1) = -6. But
p(-1) = (-1)^4 + a(-1)^2 + b(-1) - 2
p(-1) = 1 + a - b - 2
p(-1) = a - b - 1

But p(-1) = -6, so

a - b - 1 = -6
a - b = -5

That makes our two equations, two unknowns to be:

a + b = -1
a - b = -5

Solving this system of equations (which I'll omit the details) yields the result a = -3, b = 2.

5) Let q(x) = x^3 + px^2 + 6.
We are given that q(x) is exactly divisible by (x + 2) and (x - 3), so it follows that
q(-2) = 0, and q(3) = 0. But

q(-2) = (-2)^3 + p(-2)^2 + 6
q(-2) = -8 + 4p + 6
q(-2) = 4p - 2, which we now equate to 0, since q(-2) = 0.

4p - 2 = 0
p = 1/2

q(3) = 3^3 + p(3)^2 + 6
q(3) = 27 + 9p + 6
q(3) = 9p + 33, which we equate to 0, since q(3) = 0, so

9p + 33 = 0
p = -11/3

There is no mention of q anywhere.

7) x^3 + 4x^2 - 3x + 2 = (x + 1)(ax^2 + bx + c)

All we have to do is expand the right hand side.

x^3 + 4x^2 - 3x + 2 = ax^3 + bx^2 + cx + ax^2 + bx + c

Now, group like terms on the right hand side. I'll do it one step at a time to make the factoring obvious.

x^3 + 4x^2 - 3x + 2 = ax^3 + ax^2 + bx^2 + bx + cx + c

Now, factor the like terms.

x^3 + 4x^2 - 3x + 2 = ax^3 + (a + b)x^2 + (b + c)x + c

At this point, we can equate coefficients on the left and right hand sides. Notice the coefficient of x^3 on the left hand side is 1, while on the right hand side it's "a". Therefore

a = 1

Looking at the coefficient of x^2 on both sides
4 = a + b. But we know a = 1, so
4 = 1 + b. Thus,

b = 3.

Now, look at the coefficient of x; equate them componentwise.

b + c = -3. Since b = 3, then
3 + c = -3
c = -6

There is no d value. So our solutions for a, b, and c, are

a = 1, b = 3, c = -6.

8) x^3 + 4x^2 - 12x + 14 = x^3 + (mx + n)^2 + 5

Like the last question, we're going to expand the right hand side, and equate the coefficients componentwise.

x^3 + 4x^2 - 12x + 14 = x^3 + (m^2)(x^2) + 2mnx + n^2 + 5

I'll put the n^2 + 5 in brackets, just to prove that it can be compared with the constant term on the left hand side.

x^3 + 4x^2 - 12x + 14 = x^3 + (m^2)(x^2) + 2mnx + [n^2 + 5]

Equating componentwise,

4 = m^2
-12 = 2mn
14 = n^2 + 5

Since m^2 = 4, it follows that m equals two values:
m = {-2, 2}

From the second equation,
when m = 2:
-12 = 2(2)n
-12 = 4n
n = -3

when m = -2:
-12 = 2(-2)n
-12 = -4(n)
n = 3

So we have two sets of solutions:
m = 2, n = -3
m = -2, n = 3

9) x^2 = a(x + 2)^2 + b(x + 2) + c

Expand the right hand side.

x^2 = a(x^2 + 4x + 4) + bx + 2b + c
x^2 = ax^2 + 4ax + 4a + bx + 2b + c
x^2 = ax^2 + 4ax + bx + 4a + 2b + c
x^2 = ax^2 + (4a + b)x + (4a + 2b + c)

Now, we equate componentwise. Remember that we don't have coefficients for x, or a constant term, but to make this obvious,

x^2 + 0x + 0 = ax^2 + (4a + b)x + (4a + 2b + c)

This implies

a = 1
4a + b = 0
4a + 2b + c = 0

In the second equation, 4a + b = 0, but we have a = 1, so
4(1) + b = 0, implying b = 4.

Since we have a = 1 and b = 4, we use the third equation to get c:

4(1) + 2(4) + c = 0
4 + 8 + c = 0
12 + c = 0
c = -12

a = 1, b = 4, c = -12.

Answer 2

Question 5: undefined, what is q
Question 7: {a , b, c, d} = {1, 3, -6, 8}
Question 8: {m, n} = {2, -3} or { -2, 3}
Question 9: {a, b, c} = {1, -4, 4}