6.25
81.
26
I'm going to go out on a limb and assume that y2 8y c actually means y^2 + 8y + c c = 16 makes a perfect square: (y + 4)^2 = (y+4)*(y+4) = y^2 + 8y + 16
A value of the variable when the polynomial has a value of 0. Equivalently, the value of the variable when the graph of the polynomial intersects the variable axis (usually the x-axis).
A local minimum.
What value, in place of the question mark, makes the polynomial below a perfect square trinomial?x2 + 12x+ ?
None does, since there is no polynomial below.
49
49
38
64
-12
144
(b/2)^2= 64
48
There are infinitely many possible answers: c = ±4x + 33
To make the polynomial ( x^2 - 28x + ? ) a perfect square trinomial, we need to find the value that completes the square. The formula for a perfect square trinomial is ( (x - a)^2 = x^2 - 2ax + a^2 ). Here, ( a ) is half of the coefficient of ( x ), which is ( -28 ). Thus, ( a = 14 ), and we need ( a^2 = 196 ). Therefore, the value in place of the question mark is ( 196 ).