Answer: 8y4+25y3+60y2+10y+7
Step-by-step explanation:
(y2+3y+7)(8y2+y+1)
=(y2+3y+7)(8y2+y+1)
=(y2)(8y2)+(y2)(y)+(y2)(1)+(3y)(8y2)+(3y)(y)+(3y)(1)+(7)(8y2)+(7)(y)+(7)(1)
=8y4+y3+y2+24y3+3y2+3y+56y2+7y+7
=8y4+25y3+60y2+10y+7
hope this helps!:)
Given:
Seven times the smaller of two consecutive even integers is the same as -300 minus 4 times the larger integer.
To find:
The integers
.
Solution:
Let the smaller even integer be x.
So, larger even integer is x+2 because they are consecutive even integers.
Seven times the smaller integer = 7x
-300 minus 4 times the larger integer = -300-4(x+2)
Now,




Divide both sides by 11.


So, the smaller even integer is -28.

So, the larger even integer is -26.