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Find all zeros of the function f(x)=8x3−18x2−15x+25. Enter the zeros separated by commas.

 Jan 15, 2016

Best Answer 

 #2
avatar+98196 
+10

8x^3−18x^2−15x+25  = 0  

 

Since the sum of the coefficients =  0, then 1 is a root.......using synthetic divsion, we have

 

 

1  [  8     -18       -15        25  ]

                 8       -10       -25

       -------------------------------

       8     -10      -25        0

 

And the resultant polynomial set to 0 is

 

8x^2 -10x - 25 =  0       factor

 

[4x + 5] [2x - 5]  = 0

 

And each factor set to 0 produces the other two real roots :  -5/4    and  5/2

 

 

 

cool cool cool

 Jan 15, 2016
 #1
avatar
+10

Solve for x:
8 x^3-18 x^2-15 x+25 = 0

The left hand side factors into a product with three terms:
(x-1) (2 x-5) (4 x+5) = 0

Split into three equations:
x-1 = 0 or 2 x-5 = 0 or 4 x+5 = 0

Add 1 to both sides:
x = 1 or 2 x-5 = 0 or 4 x+5 = 0

Add 5 to both sides:
x = 1 or 2 x = 5 or 4 x+5 = 0

Divide both sides by 2:
x = 1 or x = 5/2 or 4 x+5 = 0

Subtract 5 from both sides:
x = 1 or x = 5/2 or 4 x = -5

Divide both sides by 4:
Answer: | x = 1,    or x = 5/2,    or x = -5/4

 Jan 15, 2016
 #2
avatar+98196 
+10
Best Answer

8x^3−18x^2−15x+25  = 0  

 

Since the sum of the coefficients =  0, then 1 is a root.......using synthetic divsion, we have

 

 

1  [  8     -18       -15        25  ]

                 8       -10       -25

       -------------------------------

       8     -10      -25        0

 

And the resultant polynomial set to 0 is

 

8x^2 -10x - 25 =  0       factor

 

[4x + 5] [2x - 5]  = 0

 

And each factor set to 0 produces the other two real roots :  -5/4    and  5/2

 

 

 

cool cool cool

CPhill Jan 15, 2016
 #3
avatar+99356 
+5

 

8x^3−18x^2−15x+25  = 0  

 

Since the sum of the coefficients =  0, then 1 is a root.......using synthetic divsion, we have

That is interesting Chris I have not ssen it expressed like that before

 

I would have used factor theory,

since  f(1) = 8-18-15+25=0 then 1 is a root which means that (x-1) is a factor.   etc

 Jan 15, 2016

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