The values for x when f(x) = 0 are:
x=-2 and x=3
My question is:
The function f(x) can be written in the form of a(x-p)(x-q)(x-r)
Write down one possible set of values for p, q, and r?
Now Ive got the correct answer x = -3, 2 and 2
Could you show me step-by step how to end up with those values?
Do you substitute p, q and r with 0 .
if roots are p,q, r, then the function is (x-p)(x-q)(x-r)
so in this case...
(x+3)(x-2)(x-2)=f(x)
To find the possible values for p, q, and r in the expression a(x-p)(x-q)(x-r), we need to use the given information that the values of x when f(x) = 0 are x = -2 and x = 3.
Step 1: Substitute the values of x into the function f(x) and set the expression equal to 0.
f(x) = a(x-p)(x-q)(x-r) = 0
Substituting x = -2:
0 = a(-2-p)(-2-q)(-2-r)
0 = a(-2-p)(-2-q)(-2-r) [equation 1]
Substituting x = 3:
0 = a(3-p)(3-q)(3-r)
0 = a(3-p)(3-q)(3-r) [equation 2]
Step 2: Expand the equations:
Expand the right side of equation 1:
0 = a(-2-p)(-2-q)(-2-r)
0 = a(4 + 2p + 2q + pq)(2 + r)
Expand the right side of equation 2:
0 = a(3-p)(3-q)(3-r)
0 = a(9 - 3p - 3q + pq)(3 - r)
Step 3: Set up a system of equations:
Now, equate the expanded equations from step 2 to 0:
a(4 + 2p + 2q + pq)(2 + r) = 0 [equation 3]
a(9 - 3p - 3q + pq)(3 - r) = 0 [equation 4]
Step 4: Solve the system of equations:
Since we have three unknowns (p, q, and r) and two equations (equations 3 and 4), we need to find values that satisfy both equations.
From equation 3, we have two options:
Option 1:
a = 0 (because if a = 0, then the entire equation will be 0)
Option 2:
4 + 2p + 2q + pq = 0 [equation 5]
2 + r = 0
From equation 4:
Option 1:
a = 0
9 - 3p - 3q + pq = 0 [equation 6]
3 - r = 0
Option 2:
9 - 3p - 3q + pq = 0 [equation 7]
3 - r = 0
Step 5: Solve equations 5, 6, and 7:
From equation 5:
4 + 2p + 2q + pq = 0
4 + (2)(-3) + (2)(-2) + (-3)(-2) = 0
4 - 6 - 4 + 6 = 0
0 = 0
From equation 6:
9 - 3p - 3q + pq = 0
9 - 3(-3) - 3(-2) + (-3)(-2) = 0
9 + 9 + 6 + 6 = 0
30 ≠ 0
From equation 7:
9 - 3p - 3q + pq = 0
9 - 3(2) - 3(2) + (2)(2) = 0
9 - 6 - 6 + 4 = 0
1 ≠ 0
So, the only option that satisfies both equations is option 1, where a = 0.
Therefore, the set of values for p, q, and r that satisfy the equation a(x-p)(x-q)(x-r) = 0 when x = -2 and x = 3 is p = -3, q = 2, and r = 2.