Find the value of the monomial 2m^2np^3 for: m=3, n=1, p=2 find the value of the monomial 2m^2np^3 for: m=0.5, n=4, p=1 rewrite as a monomial in standard form: (− 2/3 x^2y)^3*(− 3/4 xy^2)^2

Step-by-step explanation:

For the first two, fill in the given values and do the arithmetic.

... 2(3²)(1)(2³) = 2×9×8 = 144

... 2(0.5²)(4)(1³) = 2×(1/4)×4 = 2

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The outside exponents are distributed over the inside factors according to the rule:

... (a^b)^c = a^(b·c)

Then the rule ...

... (a^b)(a^c) = a^(b+c)

applies to the product.

... (− 2/3 x²y)³·(− 3/4 xy²)² = (-2/3)³x⁶y³·(-3/4)²x²y⁴

... = (-8/27)(9/16)x⁽⁶⁺²⁾y⁽³⁺⁴⁾

... = -1/6 x⁸y⁷

f(x) = x^3 -2x^2 -3x

x( x^2 -2x-3)

x( x^2 +x -3x -3)

x( x(x+1) -3(x+1))

x( x-3)( x+1)

so zeroes are 0, 3,-1

its (2) option