Let #x=4# and #y=-2#. Evaluate #(x^2-y^2(10-y^2)-:3)^2#. Apparently I have to put a question mark here ?

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Subject: Prealgebra
Focus: Squares

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Answer 1 · 2018-06-10T23:50:01

It reduces to 64

For questions of this type, we take the given values #(x=4, y=-2)# and substitute them into the expression to see what it simplifies to:

#(x^2-y^2(10-y^2)-:3)^2#

#(4^2-(-2)^2(10-(-2)^2)-:3)^2#

Now that the values are placed, we now have to work through the order of operations:

#color(red)(P)# - Parentheses (also known as Brackets)
#color(blue)(E)# - Exponents
#color(green)(M)# - Multiplication
#color(green)(D)# - Division (this has the same weight as M and so I gave it the same colour)
#color(brown)(A)# - Addition
#color(brown)(S)# - Subtraction - (again, same weight as A and so the same colour)

The bracket containing the #(10-(-2)^2)# term should be done first:

#(10-(-2)^2)#

We first square the #-2#, then subtract that result from 10:

#(10-4)=6#

#:. (4^2-(-2)^2(6)-:3)^2#

Let's now do the two squares remaining within the bracket:

# (16-4(6)-:3)^2#

Next we have the multiplication and division:

# (16-24-:3)^2#

# (16-8)^2#

We can now do the subtraction and then the square:

#8^2=64#