How to factor a common factor out of squares

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SAT Math › How to factor a common factor out of squares

Questions 1 - 7

1

According to Heron's Formula, the area of a triangle with side lengths of a, b, and c is given by the following:

Hero

where s is one-half of the triangle's perimeter.

What is the area of a triangle with side lengths of 6, 10, and 12 units?

12√5

4√14

8√14

14√2

48√77

Explanation

We can use Heron's formula to find the area of the triangle. We can let a = 6, b = 10, and c = 12.

In order to find s, we need to find one half of the perimeter. The perimeter is the sum of the lengths of the sides of the triangle.

Perimeter = a + b + c = 6 + 10 + 12 = 28

In order to find s, we must multiply the perimeter by one-half, which would give us (1/2)(28), or 14.

Now that we have a, b, c, and s, we can calculate the area using Heron's formula.

Hero

Hero2

2

Simplify the radical expression.

$\sqrt[3]{-64x^{6}y^{12}}$

$-4x^{2}y^{4}$

Explanation

$\sqrt[3]{-64x^{6}y^{12}}$

Look for perfect cubes within each term. This will allow us to factor out of the radical.

$\sqrt[3]{-64}*\sqrt[3]{x^6}*\sqrt[3]{y^{12}}$

$\sqrt[3]{-4*-4*-4}*\sqrt[3]{x^2*x^2*x^2}*\sqrt[3]{y^4*y^4*y^4}$

Simplify.

$-4x^{2}y^4$

3

Which of the following expression is equal to

$\sqrt{(45)(9)+(27)(36)}$

$9\sqrt{17}$

$9\sqrt{19}$

$7\sqrt{17}$

$9\sqrt{15}$

$7\sqrt{15}$

Explanation

$\sqrt{(45)(9)+(27)(36)}$

When simplifying a square root, consider the factors of each of its component parts:

$\sqrt{(3^2)\times5\times(3^{2})+(3^{3})(2^2)(3^2)}$

Combine like terms:

$\sqrt{5(3^4)+(2^2)(3^5)}$

Remove the common factor, :

$\sqrt{(3^4)\times(5+3\times(2^2))}$

Pull the $\sqrt{3^4}$ outside of the equation as :

$(3^2)\sqrt{5+12}=9\sqrt{17}$

4

Which of the following expressions is equal to the following expression?

$\sqrt{(27)(45)(125)}$

$225\sqrt{3}$

$125\sqrt{27}$

$75\sqrt{20}$

$135\sqrt{5}$

$205\sqrt{3}$

Explanation

$\sqrt{(27)(45)(125)}$

First, break down the component parts of the square root:

$\sqrt{(3^{3})(5\times 3^{2})(5^{3})}$

Combine like terms in a way that will let you pull some of them out from underneath the square root symbol:

$\sqrt{(5^{4})(3^4)(3)}$

Pull out the terms with even exponents and simplify:

$(5^{2})(3^{2})\sqrt{3}=225\sqrt{3}$

5

Which of the following is equal to the following expression?

$\sqrt{(16)(8)+(32)(20)}$

$2^4\sqrt{3}$

$2^{3}\sqrt{6}$

$2^{7}\sqrt{5}$

$2^{3}\sqrt{10}$

$2^{4}\sqrt{5}$

Explanation

$\sqrt{(16)(8)+(32)(20)}$

First, break down the components of the square root:

$\sqrt{(2^{4})(2^{3})+(2^{5})(2^{2})\times 5}$

Combine like terms. Remember, when multiplying exponents, add them together:

$\sqrt{(2^{7})+(2^{7})\times5}$

Factor out the common factor of :

$\sqrt{(2^{7})(1+5)}$

$\sqrt{(2^7)\times6}$

Factor the :

$\sqrt{(2^7)\times2\times3}$

Combine the factored with the :

$\sqrt{(2^{8})\times3}$

Now, you can pull $\sqrt{2^8}$ out from underneath the square root sign as :

$2^4\sqrt{3}$

6

x2 = 36

Quantity A: x

Quantity B: 6

Quantity A is greater

Quantity B is greater

The two quantities are equal

The relationship cannot be determined from the information given

Explanation

x2 = 36 -> it is important to remember that this leads to two answers.

x = 6 or x = -6.

If x = 6: A = B.

If x = -6: A < B.

Thus the relationship cannot be determined from the information given.

7

Simplify the expression.

$\sqrt{5}(2\sqrt{3}+\sqrt{12})$

$4\sqrt{15}$

$4\sqrt{30}$

$10\sqrt{3}$

$10\sqrt{15}$

$3\sqrt{2}$

Explanation

Use the distributive property for radicals.

$\sqrt{5}(2\sqrt{3}+\sqrt{12})$

Multiply all terms by $\sqrt{5}$ .

$(\sqrt{5}*2\sqrt{3})+(\sqrt{5}*\sqrt{12})$

Combine terms under radicals.

$2\sqrt{3*5}+\sqrt{12*5}$

$2\sqrt{15}+\sqrt{60}$

Look for perfect square factors under each radical. has a perfect square of . The can be factored out.

$2\sqrt{15}+\sqrt{4*15}$

$2\sqrt{15}+2\sqrt{15}$

Since both radicals are the same, we can add them.

$4\sqrt{15}$

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