Question

Can someone please help me with equating coefficients?

(x+4)(ax^2 + bx + c) = -2x^3 -7x^2 + 3x -4

Answer 1

`\large\boxed{a=-2,\ b=1,\ c=-1}`

Explanation:

`(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:}`

`\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`

Answer 2

a=-2,v=1,c=-1

EXPLANATION

`(x + 4)(a {x}^(2) + bx + c) = - 2 {x}^(3) - 7 {x}^(2) + 3x - 4`

We expand the left hand side to get,

`a {x}^(3) + b {x}^(2) + cx + 4ax^(2) + 4bx + 4c = - 2 {x}^(3) - 7 {x}^(2) + 3x - 4`

Simplify further to get:

`a {x}^(3) +(4a + b) {x}^(2) +( 4b + c)x + 4c = - 2 {x}^(3) - 7 {x}^(2) + 3x - 4`

Comparing the coefficients of the cubic terms , we have

`a {x}^(3)=-2{x}^(3)`

`a =- 2`

Comparing the quadratic terms,

`(4a + b) {x}^(2) = - 7 {x}^(2)`

`4a + b = - 7`

`4( - 2) + b = - 7`

`- 8 + b = - 7`

`b = - 7 + 8 = 1`

Comparing the constant terms,

`4c = - 4`

`c = - 1`