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?
+6248 \(f(x) = a x^3 + b x^2 + c x + d\\ f(0)= 0 \Rightarrow d = 0\\ \text{using the rest of the values we obtain 3 equations}\\ a + b + c =-5\\ -a + b -c =15\\ 8a+4b + 2c = 12\)
\(\text{adding the first two}\\ 2b = 10,~b = 5\)
\(a+c=-10\\ 8a+2c = -8\\ \text{subtract twice eq 1 from eq 2}\\ 6a = 12,~a=2\\ c= -12\)
\(f(x) = 2x^3 + 5x^2 -12x\)
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+128475 We have the form
ax^3 + bx^2 + cx + d
If f(0) = 0.....then d = 0 and we can solve this system
a(-1)^3 + b(-1)^2 + c(-1) = 15
a(1) ^3 + b(1)^2 + c(1) = - 5
a(2)^3 + b(2)^2 + c(2) = 12 simplify
-a + b - c = 15 (1)
a + b + c = -5 (2)
8a + 4b + 2c = 12 ⇒ 4a + 2b + c = 6 (3)
Add (1) and (2)
2b = 10
b = 5
Add (1) and (3)
3a + 3b = 21
a + b = 7
a + 5 = 7
a = 2
And using (1) to find c we have
-2 + 5 - c = 15
3 - c = 15
3 - 15 = c
-12 = c
So...the polynomial is
2x^3 + 5x^2 - 12x
To find the x intercepts (roots) we have
2x^3 + 5x^2 - 12x = 0 factor
x (2x^2 + 5x^2 - 12) = 0
x ( 2x - 3) (x + 4) = 0
Setting each factor to 0 and solving for x produces the x intercepts of
x = 0 x = 3/2 and x = -4
