The GCF 28a^3b and 16ab^2 A)4ab B)4a^3b C)28ab D)112a^3b^2

1 answer:

5 0

The GCF of 28 and 16 is 4 since 28/4=7 and 16/4=4 and 7 and 4 are relatively prime. The gcd of a^3 and a is a. The gcd of b and b^2 is b. Thus, the gcd of the whole thing is 4ab so A is the answer.

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Given that (x+y+z)(xy+xz+yz)=18 and that x^2(y+z)+y^2(x+z)+z^2(x+y)=6 for real numbers x, y, and z, what is the value of xyz?

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Answer:

Step-by-step explanation:

$(x+y+z)(xy+xz+yz)=18$ Equation 1

$x^2(y+z)+y^2(x+z)+z^2(x+y)=6$ Equation 2

What is the value of $xyz$ where each variable represents a real number?

Let's expand equation 1:

$(x+y+z)(xy+xz+yz)=18$

$x(xy)+x(xz)+x(yz)+y(xy)+y(xz)+y(yz)+z(xy)+z(xz)+z(yz)=18$

Simplify each term if can:

$x^2y+x^2z+xyz+y^2x+xyz+y^2z+xyz+z^2x+z^2y=18$

See if we can factor a little to get some of the left hand side of equation 2:

The first two terms have $x^2$ and if I factored $x^2$ from first two terms I would have $x^2(y+z)$ which is the first term of left hand side of equation 2.