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

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- Jun 12th 2009, 07:33 PMamath9polynomials
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 - Jun 12th 2009, 08:48 PMAmer

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

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

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

$\displaystyle 10=A+B$

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

$\displaystyle -3=-B+C$

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

$\displaystyle -3 = A -C$

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

$\displaystyle 10=A+B$

$\displaystyle -3=-B+C$

$\displaystyle -3 = A -C$

find the sum of the second and the third

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

$\displaystyle 4=2A \Rightarrow A=2 $ the rest for you - Jun 13th 2009, 02:43 AMHallsofIvy
Multiply out the right side: $\displaystyle 10x^2- 3x- 3= Ax^2+ A+ Bx^2- Bx+ Cx- C$.

$\displaystyle 10x^2- 3x- 3= (A+ B)x^2+ (C- B)x- C$

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