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THE ANSWER IS 

 (x(3)+250x(2)+100x)/((1)/(2)x(2)+25x+9) 

Final result :

2x • (x2 + 250x + 100) —————————————————————— x2 + 50x + 18

Step by step solution :

Step  1  :

1 Simplify — 2

Equation at the end of step  1  :

x)1 ————•x2)+25x)+9) (((2

Step  2  :

Equation at the end of step  2  :

)x2 ———+25x)+9) ((2

Step  3  :

Rewriting the whole as an Equivalent Fraction :

 3.1   Adding a whole to a fraction 

Rewrite the whole as a fraction using  2  as the denominator :

25x 25x • 2 25x = ——— = ——————— 1 2

Equivalent fraction : The fraction thus generated looks different but has the same value as the whole 

Common denominator : The equivalent fraction and the other fraction involved in the calculation share the same denominator

Adding fractions that have a common denominator :

 3.2       Adding up the two equivalent fractions 
Add the two equivalent fractions which now have a common denominator

Combine the numerators together, put the sum or difference over the common denominator then reduce to lowest terms if possible:

x2 + 25x • 2 x2 + 50x ———————————— = ———————— 2 2

Equation at the end of step  3  :

(x2+50x) ————————+9) (2

Step  4  :

Rewriting the whole as an Equivalent Fraction :

 4.1   Adding a whole to a fraction 

Rewrite the whole as a fraction using  2  as the denominator :

9 9 • 2 9 = — = ————— 1 2

Step  5  :

Pulling out like terms :

 5.1     Pull out like factors :

   x2 + 50x  =   x • (x + 50) 

Adding fractions that have a common denominator :

 5.2       Adding up the two equivalent fractions 

x • (x+50) + 9 • 2 x2 + 50x + 18 —————————————————— = ————————————— 2 2

Equation at the end of step  5  :

(x2+50x+18) ——————————— 2

Step  6  :

Equation at the end of step  6  :

(x2 + 50x + 18) ——————————————— 2

Step  7  :

x2+50x+18 Divide x3+250x2+100x by ————————— 2

Step  8  :

Pulling out like terms :

 8.1     Pull out like factors :

   x3 + 250x2 + 100x  =   x • (x2 + 250x + 100) 

Trying to factor by splitting the middle term

 8.2     Factoring  x2 + 250x + 100 

The first term is,  x2  its coefficient is  1 .
The middle term is,  +250x  its coefficient is  250 .
The last term, "the constant", is  +100 

Step-1 : Multiply the coefficient of the first term by the constant   1 • 100 = 100 

Step-2 : Find two factors of  100  whose sum equals the coefficient of the middle term, which is   250 .

     -100   +   -1   =   -101

     -50   +   -2   =   -52

     -25   +   -4   =   -29

     -20   +   -5   =   -25

     -10   +   -10   =   -20

     -5   +   -20   =   -25


For tidiness, printing of 12 lines which failed to find two such factors, was suppressed

Observation : No two such factors can be found !! 
Conclusion : Trinomial can not be factored

Trying to factor by splitting the middle term

 8.3     Factoring  x2+50x+18 

The first term is,  x2  its coefficient is  1 .
The middle term is,  +50x  its coefficient is  50 .
The last term, "the constant", is  +18 

Step-1 : Multiply the coefficient of the first term by the constant   1 • 18 = 18 

Step-2 : Find two factors of  18  whose sum equals the coefficient of the middle term, which is   50 .

     -18   +   -1   =   -19

     -9   +   -2   =   -11

     -6   +   -3   =   -9

     -3   +   -6   =   -9

     -2   +   -9   =   -11

     -1   +   -18   =   -19

     1   +   18   =   19

     2   +   9   =   11

     3   +   6   =   9

     6   +   3   =   9

     9   +   2   =   11

     18   +   1   =   19


Observation : No two such factors can be found !! 
Conclusion : Trinomial can not be factored

Final result :

2x • (x2 + 250x + 100) —————————————————————— x2 + 50x + 18

Nov 29, 2017
 #1
avatar+2446 
+4

If I am not mistaken, this problem requires you to generate the cubic equation of the four points listed. I see that f(0)=0, so one of the coordinates of the x-intercept is certainly located at (0,0).I also notice that there must be another zero located at \(1  because the graph crosses the x-axis there, but we cannot determine exactly where it is yet. An odd-degree polynomial cannot have an even number of real zeros (unless a multiplicity of a zero is even). We have determined already that there are 2 zeros, so a third must be lying somewhere. Yet again, we cannot determine where. 

 

Cubic functions are written in the form f(x)=ax3+bx2+cx+d. We can actually solve for d straightaway by plugging in f(0).

 

f(x)=ax3+bx2+cx+dPlug in f(0)
f(0)=a03+b02+c0+dThere is a lot of simplification that can occur here!
f(0)=dAccording to your original three points, f(0)=0, so substitute that in.
0=d 
  

 

Great! We have already solved for one of the missing variables. Our original equation is now f(x)=ax3+bx2+cx. It only gets harder from here, though. Let's plug in the remaining points, and let's see what happens!

 

f(x)=ax3+bx2+cxEvaluate this function at f(-1).
f(1)=a(1)3+b(1)2+c(1)Ok, now simplify.
f(1)=a+bcBy definition, f(1)=15.
15=a+bc 
  

 

Let's plug in the second point.

 

f(1)=a(1)3+b(1)2+c(1)Simplify from here.
f(1)=a+b+cf(1)=5, so plug that back in!
5=a+b+c 
 

 

 

Normally, I would just go right ahead and plug in the next point, but these two equations are somewhat unique. 

 

15=a+bc5=+a+b+c

 

Look at that! Both the a's and the c's will cancel out here! We can solve for b using elimination.

 

15=a+bc5=+a+b+cAdd the equations together.
10=2bDivide by 2 on both sides!
5=b 
  

 

Now, let's plug in the third point.

 

f(x)=ax3+bx2+cxEvaluate the function at f(2)
f(2)=a(2)3+b(2)2+c(2)Now, simplify.
f(2)=8a+4b+2cRemember that f(2)=12 and b=5.
12=8a+20+2cSubtract 20 from both sides.
8=8a+2cDivide by 2 because that is the GCF of the coefficients, and it will make the numbers nicer in the long run.
4=4a+c 
  

 

Ok, let's compare the previous equation with the first one we calculated. 

 

15=a+bc4=4a+c

 

It looks like more cancellation can be done, right? Well' let's do that then!

 

15=a+bc4=4a+cAdd both equations together.
11=3a+bOf course, b=5.
11=3a+5Now, solve for a.
6=3a 
2=a 
  

 

Now, plug this back into an already existing equation to find c.

 

4=4a+ca=2, so substitute that value in.
4=42+cNow, solve for c.
4=8+cSubtract 8 from both sides and isolate c.
12=c 
  

 

Ok, our function is now clearly defined for the points given. It is now f(x)=2x3+5x212x. It is time to find the remaining zeros!

 

2x3+5x212x=0Factor out a GCF first before doing anything else!
x(2x2+5x12)=0The next step involves finding factors of -24 that add up to 5. (-3 and 8). Break up the linear term into these portions.
x(2x23x+8x12)=0Now, factor by grouping. 
x[x(2x3)+4(2x3)] 
x(x+4)(2x3)=0Now, set each factor equal to zero and solve.
x=0x+4=0x=42x3=02x=3x=32=1.5

 

All the zeros have been solved for finally.
  

 

Therefore, the zeros are 4,0,1.5.

Nov 29, 2017
Nov 28, 2017
 #1
avatar+130137 
+3

 x^2 + 2x + 3

 2x^5 + 3x^4 + 8x^3 + 8x^2 + 18x + 9

 

 

We can determine this by polynomial division

 

                              2x^3    -   x^2      +    4x       + 3

x^2  +  2x  +  3  [   2x^5    +   3x^4    +    8x^3   +  8x^2  +   18x   +  9  ]

                              2x^5    +  4x^4    +    6x^3

                              __________________________________

                                          - x^4      +    2x^3     +    8x^2

                                          -x^4      -      2x^3     -     3x^2

                                        _________________________

                                                             4x^3    +  11x^2    +  18x

                                                             4x^3    +    8x^2    +  12x

                                                            __________________________

                                                                                3x^2   +  6x      +  9

                                                                                3x^2   +  6x      +  9

                                                                                _________________

                                                                                                            0

The answer is shown in red

 

 

cool cool cool

Nov 28, 2017

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