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# A Little Help?

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The polynomial f(x) has degree 3. If f(-1) = 15, f(0)= 0, f(1) = -5, and f(2) = 12, then what are the x-intercepts of the graph of f?

if someone could help me out that'dbe great, thanks!

Aug 7, 2019

#1
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One of them can be seen immediately.  What one is that?

How many (at most) can there be?

Aug 7, 2019
#2
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A 3rd degree polynomial has the  form      ax^3  + bx^2 + cx + d

Since  f(0)  = 0     then   x = 0   is one of the y intercepts  and    d  = 0

So....we can solve  for  a, b and  c     as follows :

a(-1)^3  + b(-1)^2  + c(-1) =  15  ⇒      -a   +  b - c  = 15     (1)

a(1)^3  + b(1)^2  + c(1)=  -5  ⇒              a    +  b  + c =  -5   (2)

a(2)^3  + b(2)^2  + c(2)  =  12  ⇒           8a   + 4b  + 2c  =  12    (3)

Adding  (1)  and (2)   we have that     2b  =  10 ⇒    b  = 5

Sub this into  each equation for b and we  have that

-a +5 - c  =  15 ⇒    -a  - c =  10    (4)

a +5  + c  = -5  ⇒   a  + c  = -10        (5)

8a + 4(5) + 2c =  12  ⇒   8a  + 2c = -8   (6)

Multiply (5)  by -2  and add it to (6)   and we get that     6a  =12 ⇒   a  = 2

Using (5)  to find c  we have that    2 + c  = -10 ⇒    c  =-12

So....the polynomial is   f(x)  =   2x^3 + 5x^2  - 12x      set to 0 and factor and we  get  that

(2x  - 3) (x + 4 )  = 0

Setting each factor to  0 and  solving  for  x  and  we  get the  other  two x intercepts  of   x  = 3/2    and x =  -4

See the graph here  : https://www.desmos.com/calculator/ef2xuaesrm

Aug 7, 2019