+816 When the expression \(3x^2-24x+55\) is written in the form \(a(x+d)^2+e\), where a,d , and e are constants, then what is the sum a+d+e?
+9466 3x2 - 24x + 55
Factor 3 out of the first two terms.
= 3(x2 - 8x) + 55
Add 16 and subtract 16 inside the parenthesees.
= 3(x2 - 8x + 16 - 16) + 55
Factor x2 - 8x + 16 as a perfect square trinomial.
= 3( (x - 4)2 - 16) + 55
Distribute the 3 to the terms in parenthesees.
= 3(x - 4)2 - 48 + 55
Combine -48 and 55 to get 7 .
= 3(x - 4)2 + 7
= a(x + d)2 + e
So...
a + d + e = 3 + -4 + 7 = 6
+9466 3x2 - 24x + 55
Factor 3 out of the first two terms.
= 3(x2 - 8x) + 55
Add 16 and subtract 16 inside the parenthesees.
= 3(x2 - 8x + 16 - 16) + 55
Factor x2 - 8x + 16 as a perfect square trinomial.
= 3( (x - 4)2 - 16) + 55
Distribute the 3 to the terms in parenthesees.
= 3(x - 4)2 - 48 + 55
Combine -48 and 55 to get 7 .
= 3(x - 4)2 + 7
= a(x + d)2 + e
So...
a + d + e = 3 + -4 + 7 = 6
