Find all the zeros of the function and Write the polynomial as a product of linear factors

There are no real zero

Step-by-step explanation:

this is quadraic equantion,

0 = x4 + 34x2+ 255 ( regrouping)

x2 + 25=0

x2=-25

x= ⊥5i

than

x= ⊥3i

so, the product of linaer is  f(x)= x3-7x2+11x+10

thare are no real zero also no x-intercepts

damm bro the struggle

step-by-step explanation:

d, x=8

step-by-step explanation:

simplifying

4x + -5 = 2x + 11

reorder the terms:

-5 + 4x = 2x + 11

reorder the terms:

-5 + 4x = 11 + 2x

solving

-5 + 4x = 11 + 2x

solving for variable 'x'.

move all terms containing x to the left, all other terms to the right.

add '-2x' to each side of the equation.

-5 + 4x + -2x = 11 + 2x + -2x

combine like terms: 4x + -2x = 2x

-5 + 2x = 11 + 2x + -2x

combine like terms: 2x + -2x = 0

-5 + 2x = 11 + 0

-5 + 2x = 11

add '5' to each side of the equation.

-5 + 5 + 2x = 11 + 5

combine like terms: -5 + 5 = 0

0 + 2x = 11 + 5

2x = 11 + 5

combine like terms: 11 + 5 = 16

2x = 16

divide each side by '2'.