Respuesta :
x/y=z
if we know x and z
solve for y
x/y=z
times both sides by y
x=zy
divide both sides by z
x/z=y
therefor
(x^4– 3x2 + 4x – 3)/?=(x^2+x-3)
so
[tex] \frac{x^4-3x2 + 4x-3}{x^2+x-3} [/tex]=?
solve
first factor them
[tex] \frac{(x^2+x-3)(x^2-x+1)}{x^2+x-3} [/tex]=?
x^2-x+1=?
the poly is x^2-x+1
if we know x and z
solve for y
x/y=z
times both sides by y
x=zy
divide both sides by z
x/z=y
therefor
(x^4– 3x2 + 4x – 3)/?=(x^2+x-3)
so
[tex] \frac{x^4-3x2 + 4x-3}{x^2+x-3} [/tex]=?
solve
first factor them
[tex] \frac{(x^2+x-3)(x^2-x+1)}{x^2+x-3} [/tex]=?
x^2-x+1=?
the poly is x^2-x+1
Answer:
x2 – x + 1
Step-by-step explanation:
if
(x4 – 3x2 + 4x – 3)/ p(x) =(x2 + x – 3),
then
(x4 – 3x2 + 4x – 3)/(x2 + x – 3)= p(x)
(x4 – 3x2 + 4x – 3)/(x2 + x – 3)=x2 – x + 1
to check this we can multiply
(x2 + x – 3)*(x2 – x + 1)=
(x2 + x – 3)*x2 - (x2 + x – 3)*(x) + (x2 + x – 3)*1=
x4+x3-3x2-x3-x2+3x+x2+8-3= x4 – 3x2 + 4x – 3