Finding the co-efficient of A,B and C

Question

Find the value of $A,B$ and $C$ given: $$ 3x^2 + 4 ≅ A ( x + 2)^2 + B (x + 2) + C$$ I've managed to expand the brackets, however i am still confused on what to do next. Please help and explain the process.

Answer 1

Expand both sides:

$3x^2+4 ≅ Ax^2+4Ax+4A+Bx+2B+C$

Compare coefficients of $x$.

$x^2$ gives: $3=A$

$x$ gives: $0=4A+B$

constant gives: $4=4A+2B+C$

Solve simultaneously for the 3 unknowns.

$A=3$

$0=4\times3+B$ $\to$ $B=-12$

$4=4\times3+2\times-12+C$ $\to$ $C=4+24-12=16$

Alternatively substitute in $y=x-2$ to get:

$3(y-2)^2+4≅Ay^2+By+C$

$3y^2-12y+12+4≅Ay^2+By+C$

$3y^2-12y+16≅Ay^2+By+C$

Compare coefficients of $y$ to get:

$A=3$, $B=-12$, $C=16$.

Answer 2

Expand and compare the respective power of X and X^2 and the constant term you will surely get your answer....

Answer 3

The algorithmic way uses successive Euclidean divisions, the same as the change of basis algorithm from base $10$ to base $b$ for numbers.

In detail, you divide $3x^2+4ˆ$ by $x+2$, getting a first remainder. Then you divide again the quotient by $x+2$, getting a second remainder, until the quotient is a constant. You obtain this constant as a lats remainder if you try to divide it by $x+2$.

Now the coefficients as sums of powers of $x+2$ is the list of the successive remainders, in reverse order. Furthermore,, as we divide by $x+2$, we can use Horner's method to obtain the coefficients in the divisions.

I'll take a slightly longer example to demonstrate the algorithm: $p(x)=3x^3-5x+4$, to be written as a polynomial in $x+2$: $$\begin{array}{rrrrr} &3&0&-5&4\\ &\downarrow&-6&12&-14\\ \hline \color{blue}{\times{-2}}\enspace &3 &-6&7&\color{red}{-10}\\ &\downarrow&-6&24\\ \hline \color{blue}{\times{-2}}\enspace &3 &-12& \color{red}{31}\\ &\downarrow& -6\\ \hline \color{blue}{\times{-2}}\enspace & \color{red}{3} &\color{red}{-18} \end{array}$$ Thus we have: $$3x^3-5x+4=3(x+2)^3-18(x+2)^2+31(x+2)-10.$$