Answer:
[tex]a = 2; b = 1\ and\ c = -1[/tex]
Step-by-step explanation:
Given
[tex](x+4)(ax^2+bx+c)=2x^3+9x^2+3x-4[/tex]
Required
[tex]Find\ a,b,c[/tex]
[tex](x+4)(ax^2+bx+c)=2x^3+9x^2+3x-4[/tex]
Open Bracket
[tex](x)(ax^2+bx+c)+(4)(ax^2+bx+c)=2x^3+9x^2+3x-4[/tex]
[tex]ax^3+bx^2+cx+4ax^2+4bx+4c=2x^3+9x^2+3x-4[/tex]
Collect like terms
[tex]ax^3+bx^2+4ax^2+cx+4bx+4c=2x^3+9x^2+3x-4[/tex]
By comparing coefficients; we have
[tex]ax^3=2x^3\\bx^2+4ax^2=9x^2\\cx+4bx=3x\\4c=-4[/tex]
Remove all traces of x from both sides
[tex]a=2\\b+4a=9\\c+4b=3\\4c=-4[/tex]
From the first equation;
[tex]a = 2[/tex]
From the last equation
[tex]4c = -4[/tex]
Divide both sides by 4
[tex]\frac{4c}{4} = \frac{-4}{4}[/tex]
[tex]c = \frac{-4}{4}[/tex]
[tex]c = -1[/tex]
Substitute -1 for c in the third equation
[tex]c+4b=3[/tex]
[tex]-1 + 4b = 3[/tex]
Add 1 to both sides
[tex]1-1 + 4b = 3+1[/tex]
[tex]4b = 4[/tex]
Divide both sides by 4
[tex]\frac{4b}{4} = \frac{4}{4}[/tex]
[tex]b = \frac{4}{4}[/tex]
[tex]b = 1[/tex]
Substitute 2 for a in the second equation [To confirm the value of b]
[tex]b+4(2)=9[/tex]
[tex]b + 8 = 9[/tex]
Subtract 8 from both sides
[tex]b + 8 - 8 = 9 - 8[/tex]
[tex]b = 1[/tex]
Hence;
[tex]a = 2; b = 1\ and\ c = -1[/tex]
