polynomials

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June 12th 2009, 07:33 PM
amath9

polynomials

10x^2-3x-3= A(x^2+1)+(Bx+C) (x-1)

find A B C

y=kx-5
y=6x-x^2-6

solve for k
June 12th 2009, 08:48 PM
Amer

Quote:

Originally Posted by amath9

10x^2-3x-3= A(x^2+1)+(Bx+C) (x-1)

find A B C

y=kx-5
y=6x-x^2-6

solve for k

$10x^2-3x-3 =A(x^2+1)+(Bx+C)(x-1)$

$10x^2-3x-3=Ax^2+A +Bx^2-Bx+xC-C$

the factors which multiply with $x^2$ in the right side should equal to the factor of $x^2$ in the left side so you will have

$10=A+B$

and the factors which are multiply with x in the right side should equal to the factor with x in the left side so you will have

$-3=-B+C$

and finally the terms with no x in the right side should equal to the terms with no x in the left side so you will have this

$-3 = A -C$

now you have three equation with three variables so you can solve them

$10=A+B$

$-3=-B+C$

$-3 = A -C$

find the sum of the second and the third

$-6=A-B$ find the sum of this and the first one

$4=2A \Rightarrow A=2$ the rest for you
June 13th 2009, 02:43 AM
HallsofIvy

Quote:

Originally Posted by amath9

10x^2-3x-3= A(x^2+1)+(Bx+C) (x-1)

find A B C

Multiply out the right side: $10x^2- 3x- 3= Ax^2+ A+ Bx^2- Bx+ Cx- C$ .
$10x^2- 3x- 3= (A+ B)x^2+ (C- B)x- C$
In order that these be equal for all x, corresponding coefficients must be equal: 10= A+ B, -3= C- B, and -3= -C. Solve for A, B, and C.

Quote:

y=kx-5
y=6x-x^2-6

solve for k

$y= kx- 5= 6x- x^2- 6$ . Because the left side does not have an " $x^2$ " term, there is NO value of k that will make this true for all x.