very important maths question, plz help!!?

Question

Here is the question.............i tried alot but dnt get it any1 help me plzzz ill give u extra points:

From the defination of polynomials we know tht,

P(x) = D (x) x Q(x) + R(x)

If P(x) is a cubic expression and D(x) is a linear, this means tht Q(x) is quadratic and R(x) is a constant. Hence if Q(x) = a2x^2 + a1x + a0, and R (x) = R, this means tht P(x) = D (x) x (a2x^2 + a1x + a0) + R

Suppose that polynomial P(x) = X^3 + 2x^2 -3x + 4 is divided by D (x) = x – 2, then:

(X3 + 2x2 -3x + 4) = (x – 2) x (a2x^2 + a1x + a0) + R

By expanding the brackets on the R.H.S of this equation, show tht :

a2 = 1
a1 = 2(a2) + 2
a0 = 2(a1) - 3
R = 2(a0) + 4

Answer 1

When you expand the RHS, you get
a_2 xxx + [-2a_2 + a_1] xx + [-2a_1 + a_0]x - 2a_0 + R.
This must match identically with
xxx + 2xx + -3x + 4,
and therfore

a_2 = 1.
a_1 - 2a_2 = 2 so a_1 = 4
a_0 - 2 a _1 = -3 so a_0 = 5
R - 2a_0 = 4 so R = 14.

I understand this question may be important for you understand identities, but you should go do another!

Answer 2

The remainder theorem says that

f(x)/(x-a) = Q(x) + r(x)/(x-a) ; with r(x) being the remainder.
so f(x) = (x-a)Q(x) + r(x)

Let x = a, then f(a) = r(a) where r(x) is not dependent on x

If you have a cubic expression and divide by x-2
The remainder is f(2)

The remainder of this expression is (2)³ + 2(2)² - 3(2) + 4 = 14
So R = 14.

We know that
x³ + 2x² - 3x + 4 = (x – 2) ([a2]x^2 + [a1]x + [a0]) + 14


Let x = 0, then
(0)³ + 2(0)² - 3(0) + 4 = ((0) – 2) ([a2](0)^2 + [a1](0) + [a0]) + 14
4 = -2([a0]) + 14 therefore [a0] = 5.

Let x = 1 then
(1)³ + 2(1)² - 3(1) + 4 = ((1) – 2) ([a2](1)^2 + [a1](1) + 5) + 14
1 + 2 - 3+ 4 = -([a2] + [a1] + 5) + 14
[a2] + [a1] = 5

Let x = -1 then
(-1)³ + 2(-1)² - 3(-1) + 4 = ((-1) – 2) ([a2](-1)^2 + [a1](-1) + 5) + 14
-1 + 2 + 3 + 4 = -3([a2] - [a1] + 5) + 14
8 = -3[a2] + 3[a1] - 15 + 14
9 = -3[a2] + 3[a1]
3 = [a1] - [a2]

[a1] + [a2] = 5
[a1] - [a2] = 3

Add them together:

2[a1] = 8 => [a1] = 4 and [a2] = 1.

We have [a2] = 1; [a1] = 4; [a0] = 5; R = 14 which satisfy the conditions you stated.

I hope this helps.

Answer 3

What you need to do is equate coefficients of
like powers of x on both sides of the equation.
Why? If 2 polynomials are equal, their
difference is 0, so all the coefficients of
their difference are 0.
So, a2 =1 is immediate, equating the x³ terms.
Now equate the x² terms:
a1-2a2 = 2
So a1 = 2a2 + 2.
Equate the coefficients of x:
a3 = a0-2a1
a0 = 2a1-3
Finally, equating the constant terms:
4 = -2a0+R
R = 2a0+4.
Hope that helps!

Answer 4

hi,shreya!! i think u do not need expand d RHS or apply REMAINDER THM. I hav an easy way. see in eq. that
(x-2)x (.......)??. just put x=2 in equ.it wud b..
(2)^3+2(2)^2-3(2)+4=(0)+R...
i.e., R=14..... (1)
now,put x=0& use,above result, u'll get a0=5.......(2)
now,put x=1,& use(1) & (2), u'll get a1+a2=5 .. (3)
now,put x= -1,& use(1) & (2), u'll get a1-a2=3 .. (4)
add (3) & (4), u'll get a1=4 ,a2=1.so,
R=14, a0=5, a1=4, a2=1.
EASY, ISN'T IT???/

I M ABHINAV. thanx 4 accepting my frndship. i m AN INDIAN.,

now, pls ask any Que. so dat i can talk u more by writing in ans. coloumn.and if u wanna talk me,u can go 2 my profile and add ur comment by answering 1 of my latest ques.or in same frnd que u answered on 13th.u CAN OR CANNOT ANSWER not ans. dat que, but just add ur comment in ans. coloumn. y d way, i m a boy.r u a girl or boy??. though it does not matters. sorry if dat hurt u... still, i want som points

Answer 5

I don't really get all of that other stuff, but I used synthetic division on (x^3 + 2x^2 - 3x + 4) / (x - 2) and got:

x^2 + 4x + 5 + (14 / (x - 2))

Maybe that helps?

Answer 6

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Answer 7

and just excactly why is this a very important math question?Is it like the answer to all of lifes problems if you solve it.