Question

How do you factor the trinomial `20a^2 + 47ab + 21b^2`?

Answer 1

y = (5a + 3b)(4a + 7b)

Explanation:

Use the new AC Method to factor trinomials (Socratic Search)
First, consider a as variable, and b as constant.
Solve the quadratic equation:
`y = 20a^2 + 47ab + 21 b^2 =` 10(a + p)(a + q)
Converted trinomial: `y' = a^2 + 47 ab + 420b^2 =` (a + p')(a + q').
p' and q' have same sign because ac > 0.
Factor pairs of ac = 420b^2 -->... (6b, 70b)(10b, 42b)(12b, 35b). This sum is 47b = b. Then, p' = 12b and q' = 35b.
Back to original trinomial: `p = (p')/a = (12b)/ 20 = (3b)/5` and
`q = (q')/a = (35b)/20 =( 7b)/4`.
Factored form: `y = 20(a + (3b)/5)(a + (7b)/4) = (5a + 3b)(4a + 7b)`