Can you help me with long division of this polynomial?

(7-5x^2)/(x-5)

___|_____ -5x-25
x-5|-5x^2+0x+7
___|-5x^2+25x
------------------
_________-25x+7
_________-25x+125
------------------------
______________-118

compare synthetic division
5|-5 0 7
_|_-25-125
_-5-25-118

(-5x^2+7)/(x-5) = -5x-25 -118/(x-5)

f(x)=-5x^2+7
f(5)=-5(25)+7 = -125+7=-118 yes

(7-5x²)/(x-5)
= (-5x(x-5) - 25x + 7)/(x-5)
= -5x +( -25(x-5) - 125 + 7)/(x-5)
= -5x - 25 - 118/(x-5)

Be VERY careful with your (parens)

7-5x^2/x-5 is 7 - (5x^2/x) -5 according to PEMDAS!!!
with answer: 2 -5x

This is probably what you meant:
(-5x² +7) / (x-5)
It's much different from what you wrote, AND if you put them into any math software you will get 2 different answers!

Can you help me with long division of this polynomial?
Update: 7-5x^2/x-5


Use parenthesis for the numerator and denominator please..

(7 - 5x^2)/(x - 5)
(-5x^2 + 7)/(x - 5) << I just rearranged the top in descending order of powers..

When setting up the division...make sure you have 0 as a coefficient for the missing terms here...the x term is missing so put 0x...and so..

. . . . -5x - 25
. . . .______________
x - 5)-5x^2 + 0x + 7
. . . . -5x^2 +25x
. . . . --------------------------
. . . . . . . . .-25x + 7
. . . . . . . . -25x + 125
. . . . . . . . . ------------------
. . . . . . . . . . . . . -118

The remainder is -118 and the quotient is -5x - 25 ...we can write it as..

(-5x - 25)(x - 5) - 118 or (-5x - 25) - 118/(x - 5)

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