Question

square binomials (4x^2-y^5)2

Answer 1

what is the question here ?

Answer 2

\((4x^2-y^5)^2\)

Use your rules for exponents to solve this.
\((4x^2-y^5)^2\)

When multiplying, you add exponents together, so in this case you would add it to both x and y, as well as that you will square the whole numbers.
\((4\times 4)x^{2+2}\) \(((-1)^2)\) \(y^{5+2}\)
\(16x^{2+2}+y^{5+2}\)

Then you want to add those together.
\(16x^4+y^7\)

So you would be left with the binomial.
\(16x^4+y^7\)

Answer 3

\[(4x^2-y^5)^2=(4x^2)^2+(y^5)^2-2(4x^2)(y^5)\]
\[(ax^b)^c=a^c(x)^{b \times c}\]
solve it.

Answer 4

\[=16x^4+y^{10}-8x^2y^5\]

Answer 5

so was my answer right or no

Answer 6

\[(x^a)^b=x^{a \times b}\]