divison of polynomial topic help

[bucket]

i just dont know how -6C must equal to 30

Uhm.
Well all of the first terms in the factorised equation will equal the first term in the expanded form, and all the last terms in the factorised equation must equal the last term in the expanded form, ie. the one without the pronumeral.



You can see that (which are the first terms of each of the factors) (which is the first term in the expanded form)
so then (aka -6c) must (which is the last term in expanded form)

so



now back into the equation:

then you just expand that equation to find a and b.

hope that made sense

[Mao]

riight

your question was:

If (2x-3) and (x+2) are factors of 2x^3 + ax^2 + bx + 30, find the values of a and b.

that means, when we expand , we should get the
these two are different ways of expressing the same polynomial, much like how and are the same

so when we did expand it, we have two polynomails which equals to each other
for them to be equal, the coefficients MUST match, because they are the same, like how two barcodes of the same product match exactly.
with this knowledge, we set the terms equal to each other and found the numbers

if you can understand what i said above, go back and have a look at the workings myself or enwiabe did at the start of the thread, it should make sense

[enwiabe]

Now, because we can equate co-efficients, we know that -6c MUST = 30, so c = -5.

hence f(x) = (2x - 3)(x + 2)(x - 5)

that bit i dont understand, how do we know that -6C must =30? so c = -5?

Hokay. It's clear that my skipping the 'cx + d' part is the problem. FORGET WHAT 'C' MEANT BEFORE, I'M STARTING FROM SCRATCH.

f(x) = 2x^3 + ax^2 + bx + 30

if TWO of the factors are 2x - 3 and x + 2, there exists a third linear factor... cx + d. (You need to multiply three 'x's together to get the 'x^3')
Now, if we envision the function as being factorised without knowing 'c' or 'd' it will look like:

f(x) = (2x - 3)(x + 2)(cx + d)
And we ALSO know that f(x) = 2x^3 + ax^2 + bx + 30

so now you've got to equate the two f(x)'s like so:

(2x - 3)(x + 2)(cx + d) = 2x^3 + ax^2 + bx + 30
taking the Left Hand Side ONLY and multiplying first two brackets:
(2x - 3)(x+2)(cx + d) = (2x^2 + 4x - 3x - 6)(cx + d)
                             = (2x^2 + x - 6)(cx + d), and now multiplying the remaining brackets...
                             = (2cx^3 + 2dx^2 + cx^2 + dx - 6cx - 6d)
                             = 2cx^3 + (2d + c) + (d - 6c) - 6d

Now we EQUATE the Left Hand Side with the Right Hand Side again to get:

(2x - 3)(x + 2)(cx + d) = 2x^3 + ax^2 + bx + 30

But we've expanded out the Left Hand Side, so:

2cx^3 + (2d + c) + (d - 6c) - 6d = 2x^3 + ax^2 + bx + 30

and now you equate the co-efficients on both sides to get:

2c = 2 (equation 1)
2d + c = a (equation 2)
d - 6c = b (equation 3)
-6d = 30 (equation 4)

Solve eqn 1: 2c = 2, therefore c = 1
Solve eqn 4: -6d = 30, therefore d = -5

Solve eqn 2: a = 2d + c, and if d = -5 and c = 1, then a = -10 + 1 = -9
Solve eqn 3: b = d - 6c, and i f d= -5 and c = 1, then b = -5 - 6 = -11

Therefore f(x) = 2x^3 - 9x^2 - 11x + 30

and therefore, a = -9 and b = -11!



            

[sisqo1111]

yes lol
finally i understand it
thanks so much everyone 

[cara.mel]

Hokay. It's clear that my skipping the 'cx + d' part is the problem. FORGET WHAT 'C' MEANT BEFORE, I'M STARTING FROM SCRATCH.   

I said so.

[enwiabe]

Fpppt. Knowitall