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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Thank you thank you thank you.

 

[Evergreen]

Question 1

P(x)= x^3+ax^2+bx+2

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

P(2)= 8+4a+2b+2

8+4(b-1)+2b+2=0

6b=-6
b=-1
a=b-1
a=-2​

Question 2
f(2)= 16a+8b+60+18+2=216

16a+8b=136

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

16(b-8)+8b=136

24b= 264
b=11
a=b-8
a=3

Question 3

x^3-12x^2+17x+90

by testing f(x)=0, f(5)=0 .: x-5 is a factor of P(x)

by long division

f(x)= (x-5)(x^2-7x-18)
= (x-5)(x-9)(x+2)

 

[177152]

Thank you for all your help Evergreen. I actually think I know how to do it now. Thanks

 

[Yamiyo]

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.)

4. (x+3)(x-4) is a factor
x^2-4x+3x-12
x^2-x-12
Dividing by this

                                       x^2 + 4x + 3
x^2-x-12|x^4 + 3x^3 - 13x^2 - 51x - 36
                x^4 -  x^3    -12x^2
                          4x^3  - x^2 - 51x
                          4x^3 - 4x^2 - 48x
                                      3x^2 - 3x - 36
                                      3x^2 - 3x - 36

P(x)=(x^2-x-12)(x^2+4x+3)
P(x)=(x-4)(x+1)(x+3)^2

You can get this result by dividing by (x-4), then the result of that by (x+3), but that takes longer. Other than that it's just your average division.

5. Sub y=-3x-4 into equation
-3x-4=x^3+5x^2+4x-1
x^3 + 5x^2 + 7x +3 =0
P(-3)=-27+45-21+3=0
(x+3) is a factor

                   x^2 + 2x + 1
x+3| x^3 + 5x^2 + 7x +3
        x^3  +3x^2
                   2x^2 + 7x
                   2x^2 + 6x
                                x + 3

(x+3)(x+1)^2
Solns occur when x=-3, x=-1
y=9-4 y=1-4
y=5 y=-3

(-3,5) and (-1,-3)
So basically you just sub in the line and solve using the usual long division.

6. You need to get it into the form (x-a)(x-b)(x-c) like any other degree 3 eqn.. To help you see it you can use the substitution u=sinx
Now you have 2u^3+3u^2-1=0
P(-1)=-2+3-1=0

                  2u^2 +u -1
u+1|2u^3+3u^2+ 0u-1
       2u^3 +2u^2 
                   u^2 +0u
                   u^2 + u
                            -u -1
                            -u -1

(u+1)(2u-1)(u+2)
(sinx+1)(2sinx-1)(sinx+2)
sinx=-1, 1/2 (no real solns for -2)
x=270, 30, 150

Edit: I didn't see where you said you understood it. Ah well, all the more practice for me.

 

[177152]

The other questions just gave me an idea of how to do most of them, but as for the last one I wouldn't have thought of that. I usually remember and understand the procedures well, but lately I'm struggling to heh. I've just been dividing separately; Dividing by x^2-x-12 is a bit annoying for me, I mess it up. I'll try a few like that though. I'm starting to get the hang of it though Thank you.