polynomials

• Jun 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
• Jun 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

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