Polynomials (1 Viewer)

[177152]

I'm really struggling with polynomial factorisation. It's out of the factor theorem section? Is there any other way to do them other than going through every factor of the constant? I should hope so.

Here are a few that I can't for the life of me remember how to do. Some help would be much appreciated. I would really like some basic rules and tricks for doing these sort of questions and a link to a good site? Puh puh puh-lease

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1. P(x) = x^3 + ax^2 + bx + 2 has factors of x + 1 and x - 2. Find the values of a and b.

2. The remainder, when f(x) = ax^4 + bx^3 + 15x^2 + 9x + 2 is divided by x-2
is 216, and x+1 is a factor of f(x). Find a and b.

3. Write, as a product of it's factors, x^3 - 12x^2 + 17x + 90.

4. If P(x) = x^4 + 3x^3 - 13x^2 -51x - 36 has zeros -3 and 4, write P(x) as a product of it's linear factors.

5. Find the points of intersection between the curve y = x^3 + 5x^2 + 4x - 1 and the line 3x + y + 4 = 0.

6. Solve 2sin^3x + 3sin^2x - 1 = 0 for (0<[or equal to] x <[or equal to] 360.)

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[martinb]

1.

By the factor theorem, if (x+1) and (x-2) are factors, P(-1)=0 and P(2)=0

P(-1) = -1 + a - b + 2 = 0
Therefore, a - b = -1 ...... (1)

P(2) = 8 + 4a + 2b + 2 = 0
Therefore, 2a + b = -5 ........ (2)

(1) + (2): 3a = -6
Therefore, a = -2

Sub this in (1): -2 - b = -1,
Therefore, b = -1

2.

By the remainder theorem, f(2) = 216 and f(-1) = 0

f(2) = 16a + 8b + 60 + 18 + 2 = 216
Therefore, 2a + b = 17 ............. (1)

f(-1) = a - b + 15 - 9 + 2 = 0
Therefore, a - b = -8 ............. (2)

(1) + (2): 3a = 9
Therefore, a = 3

Sub this in (2): 3 - b = -8
Therefore, b = 11

3.

Let P(x) = x^3 - 12x^2 + 17x + 90

Possible zeros of P(x) are the factors of 90.

Try P(5) = 125 - 300 + 85 + 90 = 0

Therefore, (x-5) is a factor

Now, either continue this process, use long division or by observation:

Now, P(x) = x^3 - 12x^2 + 17x + 90 = (x-5)(ax^2+bx+c), where a, b and c are real numbers

By comparing coefficients, since coefficient of x^3 term is 1, a = 1
By observation, since constant term is 90, c = 90/(-5) = -18

We now have P(x) = x^3 - 12x^2 + 17x + 90 = (x-5)(x^2+bx-18)

Since this is an identity, we can substitute values into both sides:

When x=1, 1 - 12 + 17 + 90 = (-4)(1 + b - 18)
Therefore, 96 = 68 - 4b
Therefore, b = -7

Therefore P(x) = x^3 - 12x^2 + 17x + 90 = (x-5)(x^2-7x-18)

P(x) = (x-5)(x-9)(x+2) by factorising the quadratic factor x^2+7x-18

4.

Using the method above,

P(x) = x^4 + 3x^3 - 13x^2 - 51x - 36 = (x+3)(x-4)(ax^2+bx+c)

By observation (comparing coefficients) a = 1 (since coefficient of x^4 is 1) and c = (-36)/(-12) = 3

Therefore, P(x) = x^4 + 3x^3 - 13x^2 - 51x - 36 = (x+3)(x-4)(x^2+bx+3)

When x=1, 1 + 3 - 13 - 51 - 36 = (4)(-3)(1+b+3)

Therefore, -96 = -48 - 12b
Therefore, b = 4

P(x) = x^4 + 3x^3 - 13x^2 - 51x - 36 = (x+3)(x-4)(x^2+4x+3)

Therefore, P(x) = (x+3)(x-4)(x+1)(x+3)
P(x) = [(x+3)^2](x-4)(x+1)

5.

From equation of line, y = -4 - 3x

Sub this into the curve:

-4 - 3x = x^3 + 5x^2 + 4x - 1

x^3 + 5x^2 + 7x + 3 = 0

Solve as per above examples.

When x-values have been obtained, sub back in line for coordinates.

6.

Let u = sinx

P(u) = 2u^3 + 3u^2 - 1 =0

By observation, u = -1 is a root.

Note that P'(u) = 6u^2 + 6u
Therefore P'(-1) = 0

Therefore, u = -1 is a double root

P(u) = 2u^3 + 3u^2 - 1 = [(u+1)^2](au + b)
P(u) = 2u^3 + 3u^2 - 1 = [u^2+2u+1](au + b)

By comparing coefficients, a = 2, b = -1

Therefore, P(u) = [(u+1)^2](2u - 1)

Therefore solutions of P(u) = 0 are

u = -1 and u = 1/2

But u = sinx

Therefore,

sinx = -1
x = 270 degrees

sinx = 1/2
x = 30, 150 degrees

Hope these help.