Question #43a17

2 Answers
Nov 24, 2016

Answer:

#A=2," "B=-1," "C=-2#

Explanation:

The important thing to note with identities is that it is not the same as solving an equation where terms are moved from one side to the other. In an identity, keep the terms on each side where they are, simplify each side as far as possible, and then compare what is on the left with what is on the right.

This is for the first identity.

This method involves expanding the right side and equating the coefficients of like terms.

#rArr(x+2)(Ax+B)+C#

expand the brackets using, for example, the FOIL method.

#=(Ax^2+Bx+2Ax+2B)+C#

#=color(red)(A)x^2+x(color(blue)(2A+B))+(color(magenta)(2B+C))#

#"Comparing to "color(red)( 2)x^2+color(blue)(3)xcolor(magenta)(-4)#

For these 2 expressions to be equal, then the coefficients of like terms must be equal.

#rArrA=2#

#rArr2A+B=3rArr4+B=3rArrB=-1#

#rArr2B+C=-4rArr-2+C=-4rArrC=-2#

Thus #A=2 ,B=-1" and " C=-2#

Use similar method for the other 2.

Nov 26, 2016

Answer:

See explanation

Explanation:

These appear in the process of resolving polynomial fraction

#(P_m(x))/(Q_n(x))#

into partial fractions.

The first is from

(2x^2+3x-4)/(x+2) = Ax+B+C/(x+2).

The second is from

#(5x)/((x-1)^2(2x-1)(x+3))=A/((2x-1)(x+3))+B/((x-1)(x+3))+C/((x-1)^2(2x-1))#.

The third is in a slightly different mode.

Evaluation for suitable A, B, C, .. is according to the method outlined

in the answer by Jim