SAT II Math I : Single-Variable Algebra

Study concepts, example questions & explanations for SAT II Math I

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Example Questions

Example Question #11 : Simplifying Expressions

Factor completely:

\(\displaystyle t^{3} +4t^{2}-9t -36\)

Possible Answers:

\(\displaystyle \left (t +3 \right )^{2} \left (t -4} \right )\)

\(\displaystyle \left (t ^{2}+6 \right ) \left (t-6} \right )\)

\(\displaystyle \left (t ^{2}-6 \right ) \left (t+6} \right )\)

\(\displaystyle \left (t ^{2}+9 \right ) \left (t-4} \right )\)

\(\displaystyle \left (t -3 \right ) \left (t +3 \right ) \left (t + 4} \right )\)

Correct answer:

\(\displaystyle \left (t -3 \right ) \left (t +3 \right ) \left (t + 4} \right )\)

Explanation:

The grouping technique works here:

\(\displaystyle t^{3} +4t^{2}-9t -36\)

\(\displaystyle =\left (t^{3} +4t^{2} \right )-\left (9t +36 \right )\)

\(\displaystyle =\left (t ^{2} \cdot t +t^{2} \cdot 4} \right )-\left (9 \cdot t +9 \cdot 4 \right )\)

\(\displaystyle =t ^{2} \left (t + 4} \right )-9 \left ( t + 4 \right )\)

\(\displaystyle =\left (t ^{2}-9 \right ) \left (t + 4} \right )\)

The first factor is the difference of squares and can be factored further accordingly:

\(\displaystyle \left (t ^{2}-9 \right ) \left (t + 4} \right )\)

\(\displaystyle =\left (t ^{2}- 3^{2} \right ) \left (t + 4} \right )\)

\(\displaystyle =\left (t -3 \right ) \left (t +3 \right ) \left (t + 4} \right )\)

Example Question #11 : Simplifying Expressions

Factor completely:

\(\displaystyle t^{3} - 243\)

Possible Answers:

\(\displaystyle (t - 3)(t-9) ^{2}\)

The polynomial is prime.

\(\displaystyle (t - 3)\left ( t^{2} + 81\right )\)

\(\displaystyle (t - 3)\left ( t^{2}+9t + 81\right )\)

\(\displaystyle (t - 3)\left ( t^{2}+ 3t + 81\right )\)

Correct answer:

The polynomial is prime.

Explanation:

Since the first term is a perfect cube, the factoring pattern we are looking to take advantage of is the difference of cubes pattern. However, 243 is not a perfect cube of an integer \(\displaystyle (6^{3} = 216 < 243 < 7^{3} = 343)\), so the factoring pattern cannot be applied.  No other pattern fits, so the polynomial is a prime.

Example Question #123 : Single Variable Algebra

Exponentiate:

\(\displaystyle \left (x^{2} + x + 7 \right )^{2}\)

Possible Answers:

\(\displaystyle x^{4} + x^{3} +15x^{2} + 14x+ 49\)

\(\displaystyle x^{4} + 2x^{3} +15x^{2} + 14x+ 49\)

\(\displaystyle x^{4} + 2x^{3} +8x^{2} + 14x+ 14\)

\(\displaystyle x^{4} + x^{3} +8x^{2} + 14x+ 14\)

\(\displaystyle x^{4} + x^{2} + 49\)

Correct answer:

\(\displaystyle x^{4} + 2x^{3} +15x^{2} + 14x+ 49\)

Explanation:

\(\displaystyle \left (x^{2} + x + 7 \right )^{2} = \left (x^{2} + x + 7 \right )\left (x^{2} + x + 7 \right )\)

Vertical multiplication is perhaps the easiest way to multiply trinomials.

                         \(\displaystyle x^{2} + x + 7\)

                         \(\displaystyle \underline{x^{2} + x + 7}\)

                     \(\displaystyle 7x^{2} + 7x + 49\)

              \(\displaystyle x ^{3 } + x^{2} + 7x\)

   \(\displaystyle \underline{x ^{4 } + x^{3 } + 7x^{2} \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; }\)

\(\displaystyle x^{4} + 2x^{3} +15x^{2} + 14x+ 49\)

Example Question #14 : Simplifying Expressions

Exponentiate:

 \(\displaystyle \left ( 9R - 10\right )^{3}\)

Possible Answers:

\(\displaystyle 729R^{3}-2,430R^{2} + 2,700R-1,000\)

\(\displaystyle 729R^{3}-2,430R^{2} - 2,700R+1,000\)

\(\displaystyle 729R^{3}-810R^{2} - 900R+1,000\)

\(\displaystyle 729R^{3}-1,000\)

\(\displaystyle 729R^{3}-810R^{2} + 900R-1,000\)

Correct answer:

\(\displaystyle 729R^{3}-2,430R^{2} + 2,700R-1,000\)

Explanation:

The difference of two terms can be cubed using the pattern

\(\displaystyle (A-B)^{3} = A ^{3}-3A^{2}B + 3A B^{2}-B^{3}\)

Where \(\displaystyle A = 9R, B = 10\):

\(\displaystyle (A-B)^{3} = A ^{3}-3A^{2}B + 3A B^{2}-B^{3}\)

\(\displaystyle (9R-10)^{3} = (9R) ^{3}-3(9R)^{2}\cdot 10 + 3(9R) \cdot 10^{2}-10^{3}\)

\(\displaystyle (9R-10)^{3} =729R^{3}-3\cdot 81R^{2}\cdot 10 + 3(9R) \cdot 100-1,000\)

\(\displaystyle (9R-10)^{3} =729R^{3}-2,430R^{2} + 2,700R-1,000\)

Example Question #124 : Single Variable Algebra

How many of the following are prime factors of \(\displaystyle 81y^{4}- 1\) ?

I) \(\displaystyle 9y^{2} + 1\)

II) \(\displaystyle 3y^{2} + 1\)

III) \(\displaystyle 3y+1\)

IV) \(\displaystyle 3y-1\)

Possible Answers:

Three

None

Two

One

Four

Correct answer:

Three

Explanation:

Factor \(\displaystyle y^{4}- 81\) all the way to its prime factorization.

\(\displaystyle y^{4}- 81\) can be factored as the difference of two perfect square terms as follows:

\(\displaystyle y^{4}- 81\)

\(\displaystyle =\left ( 9y^{2} \right )^{2} - 1^{2}\)

\(\displaystyle =\left (9 y^{2} + 1 \right )\left ( 9y^{2} - 1 \right )\)

\(\displaystyle 9y^{2} + 1\) is a factor, and, as the sum of squares, it is a prime. \(\displaystyle 9y^{2} - 1\) is also a factor, but it is not a prime factor - it can be factored as the difference of two perfect square terms. We continue:

\(\displaystyle \left (9 y^{2} + 1 \right )\left ( 9y^{2} - 1 \right )\)

\(\displaystyle =\left ( 9y^{2} + 1 \right )\left [ \left (3y \right ) ^{2} - 1^{2} \right ]\)

\(\displaystyle =\left ( 9y^{2} + 1 \right )\left (3y+1 \right )\left (3 y-1\right )\)

Therefore, of the given four choices, only \(\displaystyle 3y^{2} + 1\) is not a factor, so the correct response is three.

Example Question #42 : Binomials

\(\displaystyle \small \frac{x^2-5x-6}{x+1}\)

For all values \(\displaystyle x\neq -1\), which of the following is equivalent to the expression above?

Possible Answers:

\(\displaystyle \small x-2\)

\(\displaystyle \small x+6\)

\(\displaystyle \small x-3\)

\(\displaystyle \small x-6\)

Correct answer:

\(\displaystyle \small x-6\)

Explanation:

First, factor the numerator. We need factors that multiply to \(\displaystyle \small -6\) and add to \(\displaystyle \small -5\).

\(\displaystyle \small 1*-6=-6\ \text{and}\ 1+(-6)=-5\)

\(\displaystyle \small x^2-5x-6=(x-6)(x+1)\)

We can plug the factored terms into the original expression.

\(\displaystyle \small \frac{x^2-5x-6}{x+1}=\frac{(x-6)(x+1)}{(x+1)}\)

Note that \(\displaystyle \small (x+1)\) appears in both the numerator and the denominator. This allows us to cancel the terms.

\(\displaystyle \frac{(x-6)(x+1)}{(x+1)}=(x-6)\)

This is our final answer.

Example Question #16 : Simplifying Expressions

Simplify the following expression: 

\(\displaystyle 3x+18y=81\)

Possible Answers:

\(\displaystyle 3x+6y=27\)

\(\displaystyle 2x+9y=40\)

\(\displaystyle x+6y=81\)

\(\displaystyle x+18y=81\)

\(\displaystyle x+6y=27\)

Correct answer:

\(\displaystyle x+6y=27\)

Explanation:

When simplifying an equation,you must find a common factor for all values in the equation, including both sides.  

\(\displaystyle 3,18,\)and, \(\displaystyle 81\) can all be divided by \(\displaystyle 3\) so divide them all at once 

\(\displaystyle \left ( \frac{3}{3}=1, \frac{18}{3}=6, \frac{81}{3}=27\right )\).  

This leaves you with 

\(\displaystyle x+6y=27\).

Example Question #21 : Simplifying Expressions

Simplify the expression\(\displaystyle \frac{x^5(3x^2y^3)}{x^4y^2z}\)

Possible Answers:

Already in simplest form

\(\displaystyle 3x^3y\)

\(\displaystyle \frac{3x^6y}{z}\)

\(\displaystyle \frac{3x^3y}{z}\)

\(\displaystyle \frac{3x^3}{yz}\)

Correct answer:

\(\displaystyle \frac{3x^3y}{z}\)

Explanation:

\(\displaystyle \frac{x^5(3x^2y^3)}{x^4y^2z}\)

Simplify the numerator by multiplying in the \(\displaystyle \tiny x^5\) term

\(\displaystyle \frac{3x^7y^3}{x^4y^2z}\)

Cancel out like terms in the numerator and denominator.

\(\displaystyle \frac{3x^3y}{z}\)

Example Question #22 : Simplifying Expressions

Simplify:  \(\displaystyle (x^2)\cdot (\frac{x^{-1}}{x})\)

Possible Answers:

\(\displaystyle \frac{1}{x}\)

\(\displaystyle \frac{1}{x^3}\)

\(\displaystyle 1\)

\(\displaystyle \frac{1}{x^2}\)

\(\displaystyle x^3\)

Correct answer:

\(\displaystyle 1\)

Explanation:

Rewrite the denominator of the second fraction using a power.

\(\displaystyle (\frac{x^{-1}}{x}) = (\frac{x^{-1}}{x^1})\)

Using the subtraction rule of exponents, we can simplify this as one term.

\(\displaystyle x^{-1-1} = x^{-2}\)

The expression becomes:

\(\displaystyle (x^2)\cdot (x^{-2})\)

Apply the addition rule of exponents.

\(\displaystyle x^{2+(-2)} = x^0 =1\)

The answer is:  \(\displaystyle 1\)

Example Question #23 : Simplifying Expressions

\(\displaystyle \frac{(x^7 - x^6)}{ (x - 1)} =\)

Possible Answers:

\(\displaystyle x^6 - x^5\)

\(\displaystyle x^5\)

\(\displaystyle \frac{1}{x-1}\)

\(\displaystyle x^6\)

\(\displaystyle \frac{x}{(x-1)}\)

Correct answer:

\(\displaystyle x^6\)

Explanation:

Factor an \(\displaystyle x^6\) in the numerator to get:

\(\displaystyle \frac{(x^7 - x^6)}{ (x - 1)} =\frac{x^6(x-1)}{(x-1)}\)

We can now cancel out \(\displaystyle (x-1)\) from the numerator and denominator leaving the answer.

\(\displaystyle x^6\)

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