Make this a perfect square trinomial x^2+22+_ pl

Do you mean 22x? If so, the third value would be 121. You take the middle value, in this case 22, and divide it in half then square it. Basically, 22/2 = 11, then 11^2 = 121. This only works if the equation is x^2+22x+(the unknown value, which in this case would be 121). To further simplify this equation, you find a value/values that would multiply to get the third value, 121, and would also add up to the second value, 22. 11 satisfies these requirements, and since the entire equation is positive, you can rewrite it as: (x+11)^2. Hope this helps! Sorry for giving more information than necessary, just want to make sure you understand the topic!

i = 2 / - 5

step-by-step explanation:

evaluation of ( 2 - 4 ) ( p + q ) when p = 2 and q = 5 i, by putting values of p and q,                                                                    

0 = ( 2 - 4 ) ( p + q )0 = ( 2 - 4 ) ( 2 + 5 i ),   by multiplying these0 = ( 4 + 10 i - 8 - 20 i )0 = 4 - 8 + 10 i - 20 i0 = - 4 -10 i   0 = 2 ( - 2 - 5 i )0 / 2 = - 2 - 5 i- 2 - 5 i = 0- 5 i = 2i = 2 / - 5