Simplifying Expressions - SAT Math

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Question

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

For all values , which of the following is equivalent to the expression above?

Answer

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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Question

Simplify the expression.

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

Answer

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

Because we are only multiplying terms in the numerator, we can disregard the parentheses.

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

To combine like terms in the numerator, we add their exponents.

$\small \frac{x^5y^6z}{x^4y^3z^4}$

To combine like terms between the numerator and denominator, subtract the denominator exponent from the numerator exponent.

$x^{5-4}y^{6-3}z^{1-4}=xy^3z^{-3}$

Remember that any negative exponents stay in the denominator.

$\frac{xy^3}{z^3}$

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Question

Give the value of that makes the polynomial $9x^{2}+ Nx + 100$ the square of a linear binomial.

Answer

A quadratic trinomial is a perfect square if and only if takes the form

$A^{2} \pm 2AB + B^{2}$ for some values of and .

$9x^{2}+ Nx + 100 = (3x)^{2} + Nx + 10^{2}$ , so

and .

For $9x^{2}+ Nx + 100$ to be a perfect square, it must hold that

$Nx = 2AB = 2 \cdot 3x \cdot 10 = 60x$ ,

so . This is the correct choice.

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Question

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

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

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

III)

IV)

Answer

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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Question

Factor:

$27Q^{3}- 270Q^{2}+900Q -1,000$

Answer

$27Q^{3}- 270Q^{2}+900Q -1,000$

$=\left (3Q \right )^{3}- 3 \cdot 90Q^{2}+3\cdot 300Q -10^{3}$

$=\left (3Q \right )^{3}- 3 \cdot 9Q^{2}\cdot 10+3\cdot 3Q \cdot 10^{2} -10^{3}$

$=\left (3Q \right )^{3}- 3 \cdot \left (3Q \right )^{2}\cdot 10+3\cdot 3Q \cdot 10^{2} -10^{3}$

This can be factored out as the cube of a difference, where :

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

Therefore,

$27Q^{3}- 270Q^{2}+900Q -1,000$

$=\left (3Q \right )^{3}- 3 \cdot \left (3Q \right )^{2}\cdot 10+3\cdot 3Q \cdot 10^{2} -10^{3}$

$= (3Q - 10)^{3}$

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Question

Simplify:

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

Answer

To simplify, we begin by simplifying the numerator. When muliplying like bases with different exponents, their exponents are added.

For x: $\small 4+3=7$

For y: $\small 2+1=3$

For z: $\small 3+1=4$

The numerator is now .

When dividing like bases, their exponents are subtracted.

For x: $\small 7-4=3$

For y: $\small 3-(-3)=6$

For z: $\small 4-3=1$

Thus, our answer is .

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Question

Which of the following linear binomials is a factor of the polynomial $x^{3} -34x- 240$ ?

Answer

By the factor theorem, a polynomial is divisible by the linear binomial if and only if . We can use this fact to test each of the binomials by evaluating the dividend for the appropriate value of .

: Evaluate the polynomial at :

$x^{3} -34x- 240 = (-4)^{3} -34(-4)- 240 = -64 +136 - 240 = -170 eq 0$

: Evaluate the polynomial at :

$x^{3} -34x- 240 = (-6)^{3} -34(-6)- 240 = -216+204 -240= -252 eq 0$

: Evaluate the polynomial at :

$x^{3} -34x- 240 =6^{3} -34(6)- 240 = 216-204 -240= -228 eq 0$

: Evaluate the polynomial at :

$x^{3} -34x- 240 =8^{3} -34(8)- 240 = 512-272 -240= 0$

: Evaluate the polynomial at :

$x^{3} -34x- 240 =10^{3} -34(10)- 240 = 1,000-340 -240= 420$

The dividend assumes the value of 0 at , so of the choices given, is the factor.

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Question

What is increased by 40%?

Answer

A number increased by 40% is equivalent to 100% of the number plus 40% of the number. This is taking 140% of the number, or, equivalently, multiplying it by 1.4.

Therefore, increased by 40% is 1.4 times this, or

$1.4 \left (3t+ 7 \right ) = 1.4 \cdot 3t + 1.4 \cdot 7 = 4.2 t + 9.8$

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Question

Simplify the expression:

$\frac{x^2+2x-8}{x+4}$

Answer

To solve this problem, we first need to factor the numerator. We are looking for two numbers that multiply to equal -8 and sum to equal 2.

Now, we can write out our expression in fraction form.

$\frac{(x+4)(x-2)}{x+4}$

Since we have the like term in the numerator and denominator, we can cancel them out of our expression.

$\frac{(x+4)(x-2)}{x+4}= x-2$

Thus, our answer is .

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Question

Which of the following is a prime factor of $x^{6} - 1$ ?

Answer

$x^{6} - 1$ is the difference of two squares:

$x^{6} - 1 =\left ( x^{3} \right ) ^{2}- 1$

As such, it can be factored as follows:

$x^{6} - 1 =\left ( x^{3} \right ) ^{2}- 1 = \left ( x^{3} + 1 \right ) \left ( x^{3} - 1 \right )$

The first factor is the sum of cubes and the second is the difference of cubes; each can be factored further:

$x^{3} +1 = (x+1) \left ( x^{2}-x \cdot 1 + 1^{2} \right ) = (x+1) \left ( x^{2} - x + 1 \right )$

$x^{3} - 1 = (x-1) \left ( x^{2} + x \cdot 1 + 1^{2} \right ) = (x-1) \left ( x^{2} + x + 1 \right )$

Therefore,

$x^{6} - 1 = \left ( x^{3} + 1 \right ) \left ( x^{3} - 1 \right ) = (x+1) \left ( x^{2} - x + 1 \right ) (x-1) \left ( x^{2} + x + 1 \right )$

Of the choices, $x^{2} + x + 1$ appears in the prime factorization and is therefore the correct choice.

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Question

Decrease by 20%. Which of the following will this be equal to?

Answer

A number decreased by 20% is equivalent to 100% of the number minus 20% of the number. This is taking 80% of the number, or, equivalently, multiplying it by 0.8.

Therefore, decreased by 20% is 0.8 times this, or

$0.8 \left (9x + 4y \right ) = 0.8 \cdot 9x + 0.8 \cdot 4y = 7.2x + 3.2y$

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Question

Divide:

$\left ( 12 x ^{4} + 16x^{3}+ 64x^{2}+8 x \right ) \div 8x^{2}$

Answer

Divide termwise:

$\left ( 12 x ^{4} + 16x^{3}+ 64x^{2}+8 x \right ) \div 8x^{2}$

$=\frac{ 12 x ^{4} + 16x^{3}+ 64x^{2}+8 x }{ 8x^{2}}$

$=\frac{ 12 x ^{4} }{ 8x^{2}} + \frac{ 16x^{3} }{ 8x^{2}} + \frac{ 64x^{2} }{ 8x^{2}} +\frac{ 8 x }{ 8x^{2}}$

$=\frac{ 12 }{ 8 }x ^{4-2} + \frac{ 16 }{ 8}x^{3-2} + \frac{ 64 }{ 8 } +\frac{ 8 }{ 8} \cdot \frac{1}{x^{2-1}}$

$=\frac{ 3 }{ 2 }x ^{2} + 2x + 8 + \frac{1}{x}$

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Question

Exponentiate:

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

Answer

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 :

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

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

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

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

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Question

Decrease by 30%. Which of the following will this be equal to?

Answer

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, 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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Question

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

Answer

By the factor theorem, a polynomial is divisible by the linear binomial if and only if . Therefore, we want the value of that makes the polynomial equal to 0 when evaluated at .

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

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

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

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Question

Factor:

$125T^{3}+729$

Answer

$125T^{3}+729$ 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 .

$125T^{3}+729$

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

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

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

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Question

Exponentiate:

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

Answer

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

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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Question

Factor completely:

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

Answer

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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Question

Factor completely:

$t^{3} - 243$

Answer

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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Question

Simplify the following expression:

Answer

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

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

.

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