button31
contestada

Jimmy successfully factors a quadratic $4x^2 + bx + c$ as \[4x^2 + bx + c = (Ax + B)(Cx + D),\]where $A,$ $B,$ $C,$ and $D$ are integers. What are all the possible values of $A$?

Respuesta :

MathPhys

Answer:

±1, ±2, and ±4

Step-by-step explanation:

4x² + bx + c = (Ax + B) (Cx + D)

Distribute:

4x² + bx + c = ACx² + (AD + BC) x + BD

Matching the coefficients, AC = 4.  So A must be a factor of 4.  Possible values of A are therefore ±1, ±2, and ±4.

MrRoyal

The possible values of A are 1, 2 and 4

The factorized expression is given as:

[tex]4x^2 + bx + c = (Ax + B)(Cx + D)[/tex]

Open both brackets

[tex]4x^2 + bx + c = ACx^2 + ADx + BCx + BD[/tex]

By comparison, we have:

[tex]AC = 4[/tex]

[tex]AD + BC =b[/tex]

[tex]BD = c[/tex]

[tex]AC = 4[/tex] means that, the possible values of A are the factors of 4, because A and C are integers.

The factors of 4 are 1, 2 and 4

Hence, the possible values of A are 1, 2 and 4

Read more about factorized expressions at:

https://brainly.com/question/43919

Otras preguntas