ANSWER
a=-2,v=1,c=-1
EXPLANATION
[tex](x + 4)(a {x}^{2} + bx + c) = - 2 {x}^{3} - 7 {x}^{2} + 3x - 4[/tex]
We expand the left hand side to get,
[tex]a {x}^{3} + b {x}^{2} + cx + 4ax^{2} + 4bx + 4c = - 2 {x}^{3} - 7 {x}^{2} + 3x - 4[/tex]
Simplify further to get:
[tex]a {x}^{3} +(4a + b) {x}^{2} +( 4b + c)x + 4c = - 2 {x}^{3} - 7 {x}^{2} + 3x - 4[/tex]
Comparing the coefficients of the cubic terms , we have
[tex]a {x}^{3}=-2{x}^{3} [/tex]
[tex]a =- 2[/tex]
Comparing the quadratic terms,
[tex](4a + b) {x}^{2} = - 7 {x}^{2} [/tex]
[tex]4a + b = - 7[/tex]
[tex]4( - 2) + b = - 7[/tex]
[tex] - 8 + b = - 7[/tex]
[tex]b = - 7 + 8 = 1[/tex]
Comparing the constant terms,
[tex]4c = - 4[/tex]
[tex]c = - 1[/tex]
Answer:
[tex]\large\boxed{a=-2,\ b=1,\ c=-1}[/tex]
Step-by-step explanation:
[tex](x+4)(ax^2+bx+c)\qquad\text{Use FOIL}\ (a+b)(c+d)=ac+ad+bc+bd\\\\=(x)(ax^2)+(x)(bx)+(x)(c)+(4)(ax^2)+(4)(bx)+(4)(c)\\\\=ax^3+bx^2+cx+4ax^2+4bx+4c\qquad\text{combine like terms}\\\\=ax^3+(b+4a)x^2+(c+4b)x+4c\\\\ax^3+(b+4a)x^2+(c+4b)x+4c=-2x^3-7x^2+3x-4\\\\\text{Comparing coefficients of terms with the same exponents:}[/tex]
[tex]\left\{\begin{array}{ccc}a=-2&(1)\\b+4a=-7&(2)\\c+4b=3&(3)\\4c=-4&(4)\end{array}\right\\\\(4)\\4c=-4\qquad\text{divide both sides by 4}\\c=-1\\---------------------\\(2)\\\text{From (1) put}\ a=-2\\b+4(-2)=-7\\b-8=-7\qquad\text{add 8 to both sides}\\b=1\\---------------------\\(3)\\\text{From (4) put}\ c=-1\\-1+4b=3\qquad\text{add 1 to both sides}\\4b=4\qquad\text{divide both sides by 4}\\b=1[/tex]
