Question

Use the method of equating coefficients to find the values of a, b, and c: (x+4)(ax2+bx+c)=2x3+9x2+3x−4

Answer 1

`a = 2; b = 1\ and\ c = -1`

Explanation:

Given

`(x+4)(ax^2+bx+c)=2x^3+9x^2+3x-4`

Required

`Find\ a,b,c`

`(x+4)(ax^2+bx+c)=2x^3+9x^2+3x-4`

Open Bracket

`(x)(ax^2+bx+c)+(4)(ax^2+bx+c)=2x^3+9x^2+3x-4`

`ax^3+bx^2+cx+4ax^2+4bx+4c=2x^3+9x^2+3x-4`

Collect like terms

`ax^3+bx^2+4ax^2+cx+4bx+4c=2x^3+9x^2+3x-4`

By comparing coefficients; we have

`ax^3=2x^3\\bx^2+4ax^2=9x^2\\cx+4bx=3x\\4c=-4`

Remove all traces of x from both sides

`a=2\\b+4a=9\\c+4b=3\\4c=-4`

From the first equation;

`a = 2`

From the last equation

`4c = -4`

Divide both sides by 4

`(4c)/(4) = (-4)/(4)`

`c = (-4)/(4)`

`c = -1`

Substitute -1 for c in the third equation

`c+4b=3`

`-1 + 4b = 3`

Add 1 to both sides

`1-1 + 4b = 3+1`

`4b = 4`

Divide both sides by 4

`(4b)/(4) = (4)/(4)`

`b = (4)/(4)`

`b = 1`

Substitute 2 for a in the second equation [To confirm the value of b]

`b+4(2)=9`

`b + 8 = 9`

Subtract 8 from both sides

`b + 8 - 8 = 9 - 8`

`b = 1`

Hence;

`a = 2; b = 1\ and\ c = -1`