Polynomials (1 Viewer)

[wandering17]

I've been able to get the answers right by guessing the number and checking them for questions
as such:

Find the values of a,b, and c if:
a(x-1)^2 +b(x-1)+c = x^2
etc

But i can't guess this one, so is there an actually way to do this without guessing? A method or something

d) a(x+2)^2 + b(x+3)^2 + c(x+4)^2 = 2x^2 + 8x +6

 

[enigma_1]

What no

Don't use guess and check in MX1 lol. There's a method to everything.

Firstly expand the polynomial (d) and then collect all the terms in x^2 and x and just the constants.
So you add all the terms in x^2 together which equal 2x^2
Ie (a+b+c)x^2 = 2x^2
Get rid of the x^2 (divide both sides by it)
Ie (a+b+c)=2

Do the same for the terms in "x" and the terms with no "x" variable

You should get your other equations as:
4a+6b+8c=8. And

4a+9b+16c=6

From here onwards you basically use simultaneous equations for those 3 equations to find a, b and c.

 

[braintic]

Better:

Sub x=-2: b+4c=-2

Sub x=-3: a+c=0

Sub x=-4: 4a+b = 4

Then solve simultaneously.

 

[wandering17]

What no

Don't use guess and check in MX1 lol. There's a method to everything.

Firstly expand the polynomial (d) and then collect all the terms in x^2 and x and just the constants.
So you add all the terms in x^2 together which equal 2x^2
Ie (a+b+c)x^2 = 2x^2
Get rid of the x^2 (divide both sides by it)
Ie (a+b+c)=2

Do the same for the terms in "x" and the terms with no "x" variable

You should get your other equations as:
4a+6b+8c=8. And

4a+9b+16c=6

From here onwards you basically use simultaneous equations for those 3 equations to find a, b and c.

Hey thanks for your help, but what i'm stuck with is before i get to the simultaneous part, how do i add the like terms? if none of them are? Sorry if this may be a stupid question ://

Cause don't you get

ax^2+4ax+4a+bx^2+6bx+9b+cx^2 etc..?
Thanks

 

[enigma_1]

Hey thanks for your help, but what i'm stuck with is before i get to the simultaneous part, how do i add the like terms? if none of them are? Sorry if this may be a stupid question ://

Cause don't you get

ax^2+4ax+4a+bx^2+6bx+9b+cx^2 etc..?
Thanks

Yep that's correct

Basically what you do after that is you write them grouped together like:

(ax^2 +bx^2 + cx^2) + (4ax+ 6bx+ 8cx)+ (4a+ 9b+ 16c) =2x^2 + 8x+ 6

So the x^2 are together, the x are together and the constants are together

From there you just equate the coefficients pretty much

Do you get it or do you want me to write it out?

 

[enigma_1]

Also sorry, I've cleared out my inbox btw

 

[Crisium]

Yep that's correct

Basically what you do after that is you write them grouped together like:

(ax^2 +bx^2 + cx^2) + (4ax+ 6bx+ 8cx)+ (4a+ 9b+ 16c) =2x^2 + 8x+ 6

So the x^2 are together, the x are together and the constants are together

From there you just equate the coefficients pretty much

Do you get it or do you want me to write it out?

Continuing From This:

(a + b + c)x^2 + (4a + 6b + 8c)x + 4a + 9b + 16c = 2x^2 + 8x + 6

Then you compare the two sides

You can see that on the left hand side the coefficient of x^2 is a + b + c, and on the right side it is 2, so the you equate the two

a + b + c = 2 (This is your first equation)

Left hand side the coefficient of x is 4a + 6b + 8c, and on the right hand side it is 8, so you equate the two

4a + 6b + 8c = 8 (This is your second equation)

Left hand side the constant term is 4a + 9b + 16c, and on the right hand side it is 6, so you equate the two

4a + 9b + 16c = 6 (This is your third equation)

Then you solve simultaneously with all three to get the values of a, b and c.

 

[Kaido]

Guessing is some high level stuff. Beyond the MX1 course...

 

[braintic]

Guessing is some high level stuff. Beyond the MX1 course...

Nup, its really a 2 unit concept. But then, maybe you were joking ....

 

[braintic]

An easy way of doing the first example - the one OP didn't ask:

x^2
= [ (x-1) + 1 ]^2
= (x-1)^2 + 2(x-1) +1