XY=10
Y=X+3
X(X+3)=10
X^2+3X=10
X^2+3X-10
(X-2)(X+5)
The number of real zeros of the function f(x) = x3 + 4x2 + x − 6 is 3
<h3>How to determine the number of real zeros?</h3>
The equation of the function is given as:

Expand the function

Reorder the terms

Factor the expression

Factor out x -1

Expand

Factorize
](https://tex.z-dn.net/?f=f%28x%29%20%3D%20%5Bx%28x%20%2B%203%29%20%2B%202%28x%20%2B%203%29%5D%28x%20-%201%29)
Factor out x + 2

The function has been completely factored and it has 3 linear factors
Hence, the number of real zeros of the function f(x) = x3 + 4x2 + x − 6 is 3
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Answer: is x = -8
this is how i got it ______
6 x - 3 - 11 - 8 x = 2
(simplify both sides and combine like terms)
(6x + -8x ) + ( -3 + -11) = 2
- 2x + -14 = 2
-2x - 14 = 2
( then you add 14 to both sides)
- 2x - 14 = 2
+14 +14
-2x = 16
( then you divide both sides by -2)
-2x / -2 = 16/ -2
x = -8
ta da!! happy to help!
it made it here. this is for the second one.
( first, we subtract 8n from both sides)
13n + 26 = 8n - 29
-8n -8n
5n + 26 = -29
( then we subtract 26 from both sides)
5n + 26 = -29
- 26 -26
5n = - 55
(after that we divide both sides by 5, we do this to make n alone )
5n/ 5 = -55/ 5
n = -11