Math - Completing the Square Problem - Least Value

(a) Express 2x^2+12x-7 in the form a(x+b)^2+c, where the values of the constants a,b and c are to be found.

Answer: 2(X+3)^2-25

Use your answer to part (a) to find the least value of 6x^2+36x-17.

6x^2+36x-17 = 3(2(x+3)^2-25)

Is this a correct method? And if so, how do I form an answer from here. I understood this bit, but didn't understand how to get the answer in the mark scheme. They say 3c+4. I get that the least value is originally -25, so does that make the least value now -71.

How would I go about writing this as an answer? It's for WJEC board A-level Math, C1.

(a)

f(x) = 2x² + 12x - 7
. . . = 2 (x² + 6x) - 7
. . . = 2 (x² + 6x + 9) - 2(9) - 7
. . . = 2 (x + 3)² - 25

Least value of f(x) is c = -25

6x² + 36x - 17
= 6x² + 36x - 21 + 4
= 3 (2x² + 12x - 7) + 4
= 3 f(x) + 4

Least value of 6x² + 36x - 17
= 3 * least value of f(x) + 4
= 3c + 4
= 3(-25) + 4
= -71

Yes, least value is now -71

Mαthmφm

2x² + 12x - 7 = 0
2( x² + 6x) = 7
=>2( x + 3)² - 25
(c = -25)

Since 2x² + 12x = 7
Multiply both side by 3:
-> 3 ( 2x² + 12x) = 3 * 7
Next, add -17 both sides:
->3 ( 2x² + 12x) -17 = 21 -17 = 4

Least value 3c + 4 = 3(-25) + 4 = -71