Find the zero of the polynomial in each of the following cases (I) P(x) = x + 5
(II) P(x) = x - 5
(III) P(x) = 2x + 5​

(I) (-5) is a zero of P(x)

(II) 5 is a zero of P(x)

(III) (-5/2) is a zero of P(x)

Step-by-step explanation:

Here, P(x) = x + 5

To find the zeroes of P(x)

let P(x) = 0

∴ x + 5 = 0

∴ x = (-5)

Thus, (-5) is a zero of P(x)

Here, P(x) = x - 5

To find the zeroes of P(x)

let P(x) = 0

∴ x - 5 = 0

∴ x = 5

Thus, 5 is a zero of P(x)

Here, P(x) = 2x + 5

To find the zeroes of P(x)

let P(x) = 0

∴ 2x + 5 = 0

∴ 2x = -5

∴ x = (-5/2)

Thus, (-5/2) is a zero of P(x)

-TheUnknownScientist

for the friend request.

when a figure, such as triangle (xyz) is reflected over the line y = x to create x'y'z' the x-coordinates and y-coordinates of the original image (pre-image) will be reversed for the image. (e.g., (x, y) > (y, x)) this means that the pre-image position of the x- and y-coordinates switch places which in turn provides a mirror like image of the original xyz.  

i notice that when i draw a line segment from x to x', that the points are parallel from each other -- they are exactly across from each other and do not intersect with any other points.  

yes, think i would see the same characteristic if you drew the line segment connecting y with the reflecting line and then y' with the reflecting line because of the essence of reflections. because it is a mirror image, each point was effected the same way which leads me to conclude that they same would apply to y and y'.