There is no sign before 32 is given in the equation. I am showing the work with +32.
8x+32=1/5(25x−15)−4x.
By distribution property.
8x+32=x-3 Combine the like terms.
8x=x-3-32 Subtract 32 from each sides.
8x=x-35 Combine the like terms.
8x-x=-35 Subtract x from each sides.
7x=-35
Divide each sides by 7.
So, x=-5
Answer:
the one real zero is in the interval (-1, 0)
Step-by-step explanation:
Descartes' rule of signs tells you there are 0 or 2 positive real zeros. Changing the signs of the odd-degree terms and applying that rule again tells you there is one negative real zero. At the same time, those coefficients (-3, -5, -5, +7) have a negative sum, so you know ...
f(-1) = -6
f(0) = +7
so there is a zero in the interval (-1, 0).
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You can try a few values between x=0 and x=10 to see what the function does in that part of the graph. You find ...
f(1) = 10
f(2) = 21
f(3) = 58
So, it is safe to conclude that there are no real zeros for x > 0.
The only real zero of f(x) is in the interval (-1, 0).
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I like to use a graphing calculator for problems like this.
we will select each options and find zeros
(a)
for finding x-intercept , we can set g(x)=0
and then we can solve for x
now, we can factor it
we get
so, this is TRUE
(b)
we can set it to 0
and then we can solve for x
we get
so, this is FALSE
(c)
we can set it to 0
and then we can solve for x
so, this is TRUE
(d)
now, we can set it to 0
and then we can solve for x
so, this is TRUE
(e)
we have
now, we can set it to 0
and then we can solve for x
so, this is FALSE