cubic polynomails

[ben_ess12]

1. Values of x in the domain x[-5,4] where is in the set x E (a,b) find the value of b

2. given (x+a)(x+3)(x-2)=x^3+bx^2+cx+30 find a,b,c.

3. x^3+Ax-12 is exactly divisible by x+3. after finding the constant, A, use it to solve the equation x^3+Ax-12=0 to find x values.

I have no clue on any of these. any help guidance would be appreciated thanks

[xZero]

2) expand the equation and it should be straight forward from there

3) use factor theorem, if x^3+Ax-12 = 0, then when x = -3, x^3+Ax-12 will also = 0 since x+3 is a factor. Use that to find A, use long division/any other method to find the other factor and solve for x.

[monkeywantsabanana]

2. given (x+a)(x+3)(x-2)=x^3+bx^2+cx+30 find a,b,c.

expand the we get:



Gather like terms



Equate co-efficents:







Solve the equations you get:







Correct me anyone if I'm wrong.
 

3. x^3+Ax-12 is exactly divisible by x+3. after finding the constant, A, use it to solve the equation x^3+Ax-12=0 to find x values.

Therefore





Long division/short division
You should get:





Again- correct me if I'm wrong.

[ben_ess12]

thank you!! helped and i understood it

if possible i no this is stupid, could you go over how to expand a cubic quickly, hoilidays have turned my brain into mush,
and any idea for the first one?....should i jsut graph it on the calc?

[TrueTears]

1. Values of x in the domain x[-5,4] where (Image removed from quote.) is in the set x E (a,b) find the value of b

2. given (x+a)(x+3)(x-2)=x^3+bx^2+cx+30 find a,b,c.

3. x^3+Ax-12 is exactly divisible by x+3. after finding the constant, A, use it to solve the equation x^3+Ax-12=0 to find x values.

I have no clue on any of these. any help guidance would be appreciated thanks

First one:



factorise, then solve for zeros, then the cases come out nicely, no need to graph, algebra does it nicely.

[ben_ess12]

thanks true tears
i cant put the question up because it requires a graph and i dont no anywhere to do it. but i have a cubic graph, with the three turning points labelled, and the question is asking me to use the info on the graph,  find values of a,b,c,d in the equation
p(x)=ax^3+bx^2+cx+d...

any ideas :S

[brightsky]

hint: let p'(x) = 0. now use turning points (although there would only be two i think).

[xZero]

thanks true tears
i cant put the question up because it requires a graph and i dont no anywhere to do it. but i have a cubic graph, with the three turning points labelled, and the question is asking me to use the info on the graph,  find values of a,b,c,d in the equation
p(x)=ax^3+bx^2+cx+d...

any ideas :S

a cubic graph with 3 turning points?

[ben_ess12]

sorry not 3 turning points! i meant 3 points labelled on the graph :S

[jbebbo]

Are the three points the x-axis intercepts?
If they are then:
(let the x xis intercepts be 3, 5 and -6, for example)
y=m(x-3)(x-5)(x+6)
Then sub in some other point to solve for m (which accounts for any dilations)
then equate coefficients by expanding m(x-3)(x-5)(x+6) and solving
m(x-3)(x-5)(x+6)=ax^3+bx^2+cx+d

[brightsky]

even if they weren't x-intercepts, all you need to do is sub the points into the equation, and solve the three equations simultaneously.

[jbebbo]

even if they weren't x-intercepts, all you need to do is sub the points into the equation, and solve the three equations simultaneously.

true ... but I dont like simultaneous equations

[ben_ess12]

okay so
my threee points are
(0,8) (2,0) (3,8)

would my equations look like
p(=a(0)^3+b(0)^2+c(0)+d

p(0)=a(02)^3+b(2)^2+c(2)+d

p(=a(3)^3+b(3)^2+c(3)+d

i assume the y value goes where the p(x) or is that the x value?

[xZero]

it should be p(0)=a(0)^3+b(0)^2+c(0)+d, 8=a(0)^3+b(0)^2+c(0)+d etc

[ben_ess12]

me again

so i have done this with the last question above (finding the equation) and subbed in all my values and have a set of simulataneous equations looking like this

when the point (0,8) is subbed in i am left with
d=8
sub in (2,0) the equation is- 0=8a+4b+2c+8
sub in (3,8) the equatio is 8=27a+9b+3c+8

my next question is what do i do next? how do i solve 3 simulataneous equations?