Question

2r²+11r+15=0 quadratic equation by factoring​

Answer 1

Answer:

2r^2+(6+5)r+15=0

2r^2+6r+5r+15=0

2r(r+3)+5(r+3)=0

(r+3)(2r+5)=0

Answer 2

Answer:

r = - 3, r = - $\frac{5}{2}$

Step-by-step explanation:

Given

2r² + 11r + 15 = 0

Consider the factors of the product of the coefficient of the r² term and the constant term which sum to give the coefficient of the r- term.

product = 2 × 15 = 30 and sum = 11

The factors are + 6 and + 5

Use these factors to split the r- term

2r² + 6r + 5r + 15 = 0 ( factor the first/second and third/fourth terms )

2r(r + 3) + 5(r + 3) = 0 ← factor out (r + 3) from each term

(r + 3)(2r + 5) = 0

Equate each factor to zero and solve for r

r + 3 = 0 ⇒ r = - 3

2r + 5 = 0 ⇒ 2r = - 5 ⇒ r = - $\frac{5}{2}$