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SAT II Math I : Simplifying Expressions

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Example Question #11 : Simplifying Expressions

Decrease 6y + 17 by 30%. Which of the following will this be equal to?

Possible Answers:

The correct answer is not among the other responses.

1.8y +5.1

4.2y + 11.9

6y-13

4.2y + 17

Correct answer:

4.2y + 11.9

Explanation:

A number decreased by 30% is equivalent to 100% of the number minus 30% of the number. This is taking 70% of the number, or, equivalently, multiplying it by 0.7.

Therefore, 6y + 17 decreased by 30% is 0.7 times this, or

$0.7 \left (6y + 17 \right ) = 0.7 \cdot 6y +0.7 \cdot 17 = 4.2y + 11.9$

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Example Question #271 : Sat Subject Test In Math I

The polynomial $x^{4} - 27x^{2}+ 5x - A$ is divisible by the linear binomial x - 6 . Evaluate .

Possible Answers:

A =294

A = 354

A = - 354

None of the other choices gives the correct answer.

A =-294

Correct answer:

A = 354

Explanation:

By the factor theorem, a polynomial p (x) is divisible by the linear binomial x - c if and only if p(c) = 0 . Therefore, we want the value of that makes the polynomial equal to 0 when evaluated at x = 6 .

$x^{4} - 27x^{2}+ 5x - A = 0$

$6^{4} - 27 \cdot 6^{2}+ 5 \cdot 6 - A = 0$

$1,296- 27 \cdot 36+ 30- A = 0$

1,296- 972+ 30- A = 0

354- A = 0

354=A

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Example Question #271 : Sat Subject Test In Math I

Factor:

$125T^{3}+729$

Possible Answers:

$(5T+9)(5T-9)^{2}$

$(5T+9)(25T^{2}+45T +81)$

$(5T+9)(25T^{2}-45T +81)$

$(5T+9)^{3}$

$(5T+9)(25T^{2}+81)$

Correct answer:

$(5T+9)(25T^{2}-45T +81)$

Explanation:

$125T^{3}+729$ can be rewritten as $(5T)^{3} + 9^{3}$ and is therefore the sum of two cubes. As such, it can be factored using the pattern

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

where A = 5T, B = 9 .

$125T^{3}+729$

$=(5T)^{3} + 9^{3}$

$= (5T+9)[(5T)^{2}-5T \cdot 9 +9^{2}]$

$= (5T+9)(25T^{2}-45T +81)$

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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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SAT II Math I : Simplifying Expressions

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

All SAT II Math I Resources

Example Questions

Example Question #11 : Simplifying Expressions

Example Question #271 : Sat Subject Test In Math I

Example Question #271 : Sat Subject Test In Math I

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

All SAT II Math I Resources

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