How to factor a common factor out of squares

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

Questions 1 - 7

1

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 :

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

2

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:

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

3

Which real number satisfies $3^{x}*9=27^{2}$ ?

Explanation

Simplify the base of 9 and 27 in order to have a common base.

(3x)(9)=272

= (3x)(32)=(33)2

=(3x+2)=36

Therefore:

x+2=6

x=4

4

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 :

$2^4\sqrt{3}$

5

Which of the following is a factor of $4x^{4}+ 36 x^{2}$ ?

$x^{2}+ 9$

$x^{2}+ 6$

$2x^{2}+ 9$

$x^{2}+ 3$

$2x^{2}+ 3$

Explanation

The terms of $4x^{4}+ 36 x^{2}$ have $4x^{2}$ as their greatest common factor, so

$4x^{4}+ 36 x^{2} = 4x^{2} (x^{2}+9)$

$x^{2}+ 9$ is a prime polynomial.

Of the five choices, only $x^{2}+ 9$ is a factor.

6

What is,

$\frac{12}{\sqrt{18}}$ ?

$2*\sqrt{2}$

$3*\sqrt{2}$

$2*\sqrt{3}$

$3*\sqrt{3}$

$4*\sqrt{3}$

Explanation

To find an equivalency we must rationalize the denominator.

To rationalize the denominator multiply the numerator and denominator by the denominator.

$\frac{12}{\sqrt{18}}*\frac{\sqrt{18}}{\sqrt{18}}=$

$\frac{12*\sqrt{18}}{18}=$

Factor out 6,

$\frac{2*\sqrt{18}}{3}=$

Extract perfect square 9 from the square root of 18.

$\sqrt{18}=3*\sqrt{2}$

$\frac{2*3*\sqrt{2}}{3}=$

$2*\sqrt{2}$

7

Simplify $\frac{8^{3} \times 3}{2^{6}\times 27}$

$\frac{8}{9}$

$\frac{9}{8}$

$\frac{16}{9}$

$\frac{4}{3}$

$\frac{32}{9}$

Explanation

The easiest way to approach this problem is to break everything into exponents. $8^{3}$ is equal to $2^{9}$ and 27 is equal to $3^{3}$ . Therefore, the expression can be broken down into $\frac{2^{9} \times 3}{2^{6}\times 3^{3}}$ . When you cancel out all the terms, you get $\frac{2^{3}}{3^{2}}$ , which equals $\frac{8}{9}$ .

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