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Start by factoring each expression.
(x-5)(x-4)(x+6)
Substitute values
The solutions for this equation are x= 9 ± 12. Let's separate them into the positive and negative cases.
| x=9 ± 1/2 | |
|---|---|
| x_1=9+1/2 | x_2=9-1/2 |
| x_1=10/2 | x_2=8/2 |
| x_1=5 | x_2=4 |
Substitute values
1^a=1
Identity Property of Multiplication
- a(- b)=a* b
Add terms
Calculate root
| x=- 1 ± 11/2 | |
|---|---|
| x_1=- 1+11/2 | x_2=- 1-11/2 |
| x_1=10/2 | x_2=- 12/2 |
| x_1=5 | x_2=- 6 |
Knowing that x= 5 and x= - 6 are the zeros of the trinomial, we can write the factored form of the quadratic expression. (x- 5)(x-( - 6)) ⇔ (x-5)(x+6) We can now write both expressions as the product of their prime factors. x^2-9x+20 &= (x-5) (x-4) x^2+x-30 &= (x-5) (x+6) Finally, we write the product of the prime factors, each raised to the greatest power that occurs in their expression. (x-5) (x-4) (x+6)