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SAT II Math I : Single-Variable Algebra

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Example Question #121 : Single Variable Algebra

Factor completely:

$t^{3} +4t^{2}-9t -36$

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

The grouping technique works here:

$t^{3} +4t^{2}-9t -36$

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

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

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

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

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

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

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

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

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Example Question #271 : Sat Subject Test In Math I

Factor completely:

$t^{3} - 243$

Possible Answers:

$(t - 3)(t-9) ^{2}$

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

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

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

The polynomial is prime.

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 $(6^{3} = 216 < 243 < 7^{3} = 343)$ , so the factoring pattern cannot be applied. No other pattern fits, so the polynomial is a prime.

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Example Question #122 : Single Variable Algebra

Exponentiate:

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

Possible Answers:

$x^{4} + 2x^{3} +15x^{2} + 14x+ 49$

$x^{4} + x^{3} +15x^{2} + 14x+ 49$

$x^{4} + x^{2} + 49$

$x^{4} + x^{3} +8x^{2} + 14x+ 14$

$x^{4} + 2x^{3} +8x^{2} + 14x+ 14$

Correct answer:

$x^{4} + 2x^{3} +15x^{2} + 14x+ 49$

Explanation:

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

Vertical multiplication is perhaps the easiest way to multiply trinomials.

$x^{2} + x + 7$

$\underline{x^{2} + x + 7}$

$7x^{2} + 7x + 49$

$x ^{3 } + x^{2} + 7x$

$\underline{x ^{4 } + x^{3 } + 7x^{2} \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; }$

$x^{4} + 2x^{3} +15x^{2} + 14x+ 49$

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Example Question #121 : Single Variable Algebra

Exponentiate:

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

Possible Answers:

$729R^{3}-2,430R^{2} + 2,700R-1,000$

$729R^{3}-810R^{2} - 900R+1,000$

$729R^{3}-1,000$

$729R^{3}-810R^{2} + 900R-1,000$

$729R^{3}-2,430R^{2} - 2,700R+1,000$

Correct answer:

$729R^{3}-2,430R^{2} + 2,700R-1,000$

Explanation:

The difference of two terms can be cubed using the pattern

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

Where A = 9R, B = 10 :

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

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

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

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

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Example Question #122 : Single Variable Algebra

How many of the following are prime factors of $81y^{4}- 1$ ?

I) $9y^{2} + 1$

II) $3y^{2} + 1$

III) 3y+1

IV) 3y-1

Possible Answers:

Three

Four

One

Two

None

Correct answer:

Three

Explanation:

Factor $y^{4}- 81$ all the way to its prime factorization.

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

$y^{4}- 81$

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

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

$9y^{2} + 1$ is a factor, and, as the sum of squares, it is a prime. $9y^{2} - 1$ 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:

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

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

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

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

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Example Question #1 : How To Find The Solution Of A Rational Equation With A Binomial Denominator

$\small \frac{x^2-5x-6}{x+1}$

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

Possible Answers:

$\small x-3$

$\small x+6$

$\small x-2$

$\small x-6$

Correct answer:

$\small x-6$

Explanation:

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

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

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

We can plug the factored terms into the original expression.

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

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

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

This is our final answer.

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Example Question #126 : Single Variable Algebra

Simplify the following expression:

3x+18y=81

Possible Answers:

3x+6y=27

x+18y=81

2x+9y=40

x+6y=81

x+6y=27

Correct answer:

x+6y=27

Explanation:

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

3,18, and, can all be divided by so divide them all at once

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

This leaves you with

x+6y=27 .

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Example Question #21 : Simplifying Expressions

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

Possible Answers:

$\frac{3x^3y}{z}$

3x^3y

Already in simplest form

$\frac{3x^6y}{z}$

$\frac{3x^3}{yz}$

Correct answer:

$\frac{3x^3y}{z}$

Explanation:

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

Simplify the numerator by multiplying in the $\tiny x^5$ term

$\frac{3x^7y^3}{x^4y^2z}$

Cancel out like terms in the numerator and denominator.

$\frac{3x^3y}{z}$

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Example Question #122 : Single Variable Algebra

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

Possible Answers:

$\frac{1}{x^3}$

$\frac{1}{x^2}$

x^3

$\frac{1}{x}$

Correct answer:

Explanation:

Rewrite the denominator of the second fraction using a power.

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

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

$x^{-1-1} = x^{-2}$

The expression becomes:

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

Apply the addition rule of exponents.

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

The answer is:

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Example Question #21 : Simplifying Expressions

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

Possible Answers:

$\frac{1}{x-1}$

x^6

x^5

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

x^6 - x^5

Correct answer:

x^6

Explanation:

Factor an x^6 in the numerator to get:

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

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

x^6

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SAT II Math I : Single-Variable Algebra

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

All SAT II Math I Resources

Example Questions

Example Question #121 : Single Variable Algebra

Example Question #271 : Sat Subject Test In Math I

Example Question #122 : Single Variable Algebra

Example Question #121 : Single Variable Algebra

Example Question #122 : Single Variable Algebra

Example Question #1 : How To Find The Solution Of A Rational Equation With A Binomial Denominator

Example Question #126 : Single Variable Algebra

Example Question #21 : Simplifying Expressions

Example Question #122 : Single Variable Algebra

Example Question #21 : Simplifying Expressions

All SAT II Math I Resources

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