Question

Factoring the expression 20a^3b^3 – 24a^5b^2 + 4a^3b^2 gives a new expression of the formUa^xb^y (Wa^2 + Vb+ z), where U > 0.What is the value of U?What is the value of W?What is the value of V?What is the value of Z?What is the value of x?What is the value of y?

Answer 1

Answer:

The value of U is 4.

The value of W is -6.

The value of V is 5.

The value of Z is 1.

The value of x is 3

The value of y is 2.

Step-by-step explanation:

Let's equal both expressions:

$20a^3b^3-24a^5b^2+4a^3b^2=Ua^xb^y(Wa^2+Vb+Z)$

Now we factorize the expression on the left-side of the equality, finding the common terms.

Between 20, 24, 4.

Divisors of 20 = {1,2,4,5,10,20}

Divisors of 24 = {1,2,3,4,6,8,12,24}

Divisors of 4 = {1,2,4}

Greatest common divisor: 4, which is the common term.

Between a^3,a^5,a^3

The one with the lowest exponent, which is a^3

Between b^3,b^2,b^2

Lowest exponent is b^2

Common term: 4*a³*b²

So the expression can be rewritten as:

$4a^3b^2(\frac{20a^3b^3}{4a^3b^2}-\frac{24a^5b^2}{4a^3b^2}+\frac{4a^3b^2}{4a^3b^2})=Ua^xb^y(Wa^2+Vb+Z)$

Now we solve the divisions:

$\frac{20a^3b^3}{4a^3b^2}=\frac{20}{4}\ast\frac{a^3}{a^3}\ast\frac{b^3}{b^2}=5\ast b^{3-2}=5b$$\frac{24a^5b^2}{4a^3b^2}=\frac{24}{4}\ast\frac{a^5}{a^3}\ast\frac{b^2}{b^2}=6\ast a^{5-3}=6a^2$$\frac{4a^3b^2}{4a^3b^2}=1$

Replacing:

$4a^3b^2(5b-6a^{^2}+1)=Ua^xb^y(Wa^2+Vb+Z)$

Just a small adjustment for formatting

$4a^3b^2(-6a^2+5b+1)=Ua^xb^y(Wa^2+Vb+Z)$

Now, comparing the left side of the equality with the right side.

U = 4, x = 3, y = 2, W = -6, V = 5, Z = 1.

The value of U is 4.

The value of W is -6.

The value of V is 5.

The value of Z is 1.

The value of x is 3

The value of y is 2.