Use factor theorem to find 3 solutions of a cubic equation

Question

Here is the question:

If the 3 solutions fo a cubic function are r1,r2,r3 use the factor theorem to write the equation of the polynomial in standard form. Make sure to combine all like terms.

Please Show work so I can understand the solution better.

Answer 1

Since the three solutions are r₁, r₂, and r₃, the three factors of the cubic equation will be (x-r₁), (x-r₂), and (x-r₃). To write the cubic equation, we need to multiply these factors together:

1. f(x)=(x-r₁)(x-r₂)(x-r₃)
2. FOIL the first two factors: f(x)= (x²-r₂x-r₁x+r₁r₂)(x-r₃)
3. Distribute each term of the first polynomial to each term of the second binomial: f(x)= x³-r₃x²-r₂x²+r₂r₃x-r₁x²+r₁r₃x+r₁r₂x-r₁r₂r₃
4. These subscripts and exponents can look confusing, so let’s rearrange the terms by degree. This will make it easier to combine like terms: f(x)=x³-r₁x²-r₂x²-r₃x²+r₁r₃x+r₂r₃x+r₁r₂x-r₁r₂r₃
5. Now look at each degree as a group to combine like terms: f(x)=x³+(-r₁x²-r₂x²-r₃x²)+(r₁r₃x+r₂r₃x+r₁r₂x)-r₁r₂r₃
6. By factoring our the GCF of both groups in parentheses, we can combine like terms: f(x)=x³-(r₁+r₂+r₃)x²+(r₁r₃+r₂r₃+r₁r₂)x-r₁r₂r₃

Let’s choose some numbers to see how this works:

Let r₁=1, r₂=2, and r₃=3. If we plug those numbers into the final answer from step 6, we get: f(x)=x³-(1+2+3)x²+((1)(3)+(2)(3)+(1)(2))x-(1)(2)(3)

This would simplify to: f(x)=x³-6x²+11x-6.

Lets see if this works: If we know that the three solutions are x=1,2, and 3, then the three factors are (x-1), (x-2), and (x-3). FOIL the first two factors to get: (x²-3x+2). Multiply that trinomial with (x-3) to get: x³-3x²-3x²+9x+2x-6. Combine like terms to get: x³-6x²+11x²-6.

Answer 2

f(x) = (x - r₁)(x - r₂)(x - r₃)

f(x) = (x² - xr₁ - xr₂ + r₁r₂)(x - r₃)

f(x = x³ - x²r₁ - x²r₂ + xr₁r₂ - x²r₃ + 3xr₁ + 3xr₂ - 3r₁r₂r₃