Alg 2 Help Please!!! functions & factoring & table of values?

Question

Consider the function f(x) = x^3 + 2x^2 - 25x - 50 .

a. Find the possible rational roots of f(x) = 0.

b. Use synthetic division to divide f(x) by x + 5. Show all your work. Use your answer to explain whether or not x + 5 is a factor of f(x).

c. Factor f(x) completely. Show all your work.

d. Sketch the graph of f(x) by plotting the x-intercepts, finding and plotting additional points on the graph, and using what you know about the graph’s end behavior. (hint: count by 10’s on your y axis)

For d, just make a table of values.

Answer 1

Given that,

                  f(x) = x^3 + 2x^2 - 25x - 50

a)  Let f(x) = 0

        => x^3 + 2x^2 - 25x - 50 = 0

        => x^2(x + 2) -25(x +2) = 0      

        => (x^2 - 25) (x + 2) = 0          

        => (x^2 - 5^2) (x + 2) = 0          [(x^2 - 5^2) is in the form a^2 - b^2 which is equal to (a+b)(a-b)]

        => (x + 5)(x - 5)(x +2) = 0         [Here (x + 5), (x - 5) and (x +2) are factors of f(x)]

        => x + 5 = 0, x - 5 = 0, x + 2 =0

        => x = -5, x = 5, x = -2

        => x = -5, 5, -2

Therefore the roots of f(x) are -2, 5 and -5.

Answer 2

b) Given that f(x) = x^3 + 2x^2 - 25x - 50

     Dividing f(x) by x+5 using synthetic division,

                                   -5  |  1    2    -25   -50

                                         |  0   -5     15    50

                                         |____________________

                                             1   -3    -10    0      => x^2 -3x -10 = 0

    We get the last value zero hence it is proved that (x+5) is the factor of f(x).   

                                                                   

Answer 3

c) Given that,

                  f(x) = x^3 + 2x^2 - 25x - 50

        Let f(x) = 0

        => x^3 + 2x^2 - 25x - 50 = 0

        => x^2(x + 2) -25(x +2) = 0      

        => (x^2 - 25) (x + 2) = 0          

        => (x^2 - 5^2) (x + 2) = 0          [(x^2 - 5^2) is in the form a^2 - b^2 which is equal to (a+b)(a-b)]

        => (x + 5)(x - 5)(x +2) = 0       

  Therefore (x + 5), (x - 5) and (x +2) are factors of f(x).

Answer 4

d) The graph of f(x) can be as follows:

                              

From the graph the additional points are (5.3,20), (-3,6.19), (0,-50), (-6.4,-70) and (2.3,-85).

Answer 5

d) The polynomial function f (x ) =  x ³ + 2x ²- 25x - 50

y = (x + 2) (x - 5)(x + 5)

1) Zeros y = (x + 2) (x - 5)(x + 5) using zeros to graph the polynomial.

Real zeros are x  intercepts of the graph.

To find x intercepts substitute y = 0 in the function.

(x + 2) (x - 5) (x + 5) = 0

Apply zero product property.

x + 2 = 0 , x - 5 = 0 and x  + 5 = 0

x = -2 , x = 5 and  x = -5.

x intercepts are (0, -2), (0, 5) and (0, -5).

2)Test points

Make the table of values for the polnomial.

Here test 4 points to determine whether the graph of polynomials lies above or below the x axis.

Choose values for x and find the corresponding values for y.

x	y = (x + 2) (x - 5)(x + 5)	(x, y )
-6	y = (-6 + 2) (-6 - 5)(-6 + 5) = -44	(- 6, -44)
0	y  = (0 + 2) (0 - 5)(0 + 5) = -50	(0, -50)
6	y = (6 + 2) (6 - 5)(6 + 5) = 88	(6, 88)
2	y = (2 + 2) (2 - 5)(2 + 5) = -84	(2, -84)

3) End behavior y =  x ³ + 2x ²- 25x - 50

Degree of the polynomial is 3 and leading coefficient 1.

The graph of a polynomial function is always a smooth curve; that is, it has no breaks or corners.

All odd degree polynomials behave on thier ends like cubics.

All odd degree polynomials  have ends that head off in opposite directions.depending on whether the polynomial has, respectively, a positive or negative leading coefficient.

The above polynomial odd degree  polynomial with a positive leading coefficient .

So the graph falls to the left and rises to the right.

4)Graph

1.Draw a coordinate plane.

2.Plot the x  intercepts and coordinate points found in the table.

3.Then sketch the graph, connecting the points with a smooth curve.

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