Respuesta :
The zeros of the polynomial are 1, 2, -2 and -3, so this polynomial must have at least one of each of these factors:
(x-1), (x-2), (x-(-2)), and (x-(-3)); rewriting: (x-1), (x-2), (x+2), and (x+3).
Thus, any such polynomial must have a factor (x-1)(x-2)(x+2)(x+3).
The simplest such polynomial we can think of, is p(x)=(x-1)(x-2)(x+2)(x+3).
To write in standard form, lets first multiply the factors two by two as follows:
[tex](x-2)(x+2)=x^2-4[/tex] by the difference of squares formula,
[tex](x-1)(x+3)=x^2+2x-3[/tex].
Next, we multiply our results:
[tex](x^2-4)(x^2+2x-3)=x^2(x^2+2x-3)-4(x^2+2x-3)[/tex]
[tex]=x^4+2x^3-3x^2-4x^2-8x+12=x^4+2x^3-7x^2-8x+12[/tex].
Answer: [tex]x^4+2x^3-7x^2-8x+12[/tex]
(x-1), (x-2), (x-(-2)), and (x-(-3)); rewriting: (x-1), (x-2), (x+2), and (x+3).
Thus, any such polynomial must have a factor (x-1)(x-2)(x+2)(x+3).
The simplest such polynomial we can think of, is p(x)=(x-1)(x-2)(x+2)(x+3).
To write in standard form, lets first multiply the factors two by two as follows:
[tex](x-2)(x+2)=x^2-4[/tex] by the difference of squares formula,
[tex](x-1)(x+3)=x^2+2x-3[/tex].
Next, we multiply our results:
[tex](x^2-4)(x^2+2x-3)=x^2(x^2+2x-3)-4(x^2+2x-3)[/tex]
[tex]=x^4+2x^3-3x^2-4x^2-8x+12=x^4+2x^3-7x^2-8x+12[/tex].
Answer: [tex]x^4+2x^3-7x^2-8x+12[/tex]
Answer:
What do you mean it's not an option:
Step-by-step explanation:
It's option B on Connexus (All credit to the brainy dude down bellow)
