ISEE Upper Level (grades 9-12) Mathematics Achievement - ISEE Upper Level Mathematics Achievement

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Question

Rewrite the polynomial in standard form:

$9x^{4} - 3x ^{2} + 8x^{5} -2x + 7 - 4x^{3}$

Answer

The degree of a term of a polynomial with one variable is the exponent of that variable. The terms of a polynomial in standard form are written in descending order of degree. Therefore, we rearrange the terms by their exponent, from 5 down to 0, noting that we can rewrite the and constant terms with exponents 1 and 0, respectively:

$9x^{4} - 3x ^{2} + 8x^{5} -2x + 7 - 4x^{3}$

$= 9x^{4} - 3x ^{2} + 8x^{5} -2x ^{1} + 7x ^{0} - 4x^{3}$

$=8x^{5} + 9x^{4}- 4x^{3} - 3x ^{2} -2x ^{1}+ 7x^{0}$

$=8x^{5} + 9x^{4}- 4x^{3} - 3x ^{2} -2x+ 7$

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Question

Simplify:

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

Answer

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

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

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

$=\frac{4x^{2}y^{2}}{x^{4}y^{4}}$

$=\frac{4}{x^{2}y^{2}}$

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Question

Assume that . Simplify:

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

Answer

$\frac{(x^{2}+y^{2})^{2} + (x^{2} + y^{2})^{2}}{x^{4}y^{4}}$

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

$=\frac{x^{4}+x^{4}+2x^{2}y^{2}-2x^{2}y^{2}+y^{4}+y^{4} }{x^{4}y^{4}}$

$=\frac{2x^{4}+2y^{4} }{x^{4}y^{4}}$

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Question

If , simplify:

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

Answer

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

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

$=\frac{(x^2+2xy+y^2)}{x+y}$

$=\frac{(x+y)^2}{x+y}=x+y$

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Question

If , simplify:

$\frac{(x+\frac{1}{x})^3+(x-\frac{1}{x})^3}{\frac{2}{x}}$

Answer

$\frac{(x+\frac{1}{x})^3+(x-\frac{1}{x})^3}{\frac{2}{x}}$

$=\frac{(x^3+3x^2\times \frac{1}{x}+3x\times \frac{1}{x^2}+\frac{1}{x^3})+(x^3-3x^2\times \frac{1}{x}+3x\times \frac{1}{x^2}-\frac{1}{x^3})}{\frac{2}{x}}$

$=(x^3+x^3)+(3x-3x)+(\frac{3}{x}+\frac{3}{x})+(\frac{1}{x^3}-\frac{1}{x^3})$

$=\frac{2x^3+\frac{6}{x}}{\frac{2}{x}}$

$=\frac{\frac{2x^4+6}{x}}{\frac{2}{x}}$

$=\frac{2x^4+6}{2}$

$=\frac{2(x^4+3)}{2}$

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Question

Define

What is $g(x ^{2} + 3x)$ ?

Answer

Substitute $x ^{2} + 3x$ for in the definition:

$g(x ^{2} + 3x ) = 2 \left ( x ^{2} + 3x \right ) - 5$

$g(x ^{2} + 3x ) = 2 \left ( x ^{2}\right ) + 2 \left ( 3x \right ) - 5$

$g(x ^{2} + 3x ) = 2 x ^{2} + 6x - 5$

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Question

Add:

Answer

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Question

Simplify:

Answer

Start by reordering the expression to group like-terms together.

Combine like-terms to simplify.

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Question

Simplify:

Answer

We can expand the first term using FOIL:

Reorder the expression to group like-terms together.

Simplify by combining like-terms.

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Question

Simplify:

Answer

Expand each term by using FOIL:

Rearrange to group like-terms together.

Simplify by combining like-terms.

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Question

Simplify the expression:

$6x^{4} + 8x^{3} - 4x^{2}$

Answer

$6x^{4} + 8x^{3} - 4x^{2}$

$=2x^{2}(3x^{2}+4x-2)$

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Question

Simplify:

Answer

In order to add exponential terms, both the base and the exponent must be the same. So we can write:

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Question

Evaluate:

$\frac{x^0+y^0+1}{x^0-y^0-1}$

Answer

Based on the zero-exponent rule we have:

That means any non-zero number raised to the zero power is equal to . So we can write:

$\frac{1+1+1}{1-1-1}=\frac{3}{-1}=-3$

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Question

Which of the following is equivalent to the expression below?

$4^{x}\cdot2^{5}$

Answer

When exponents are multiplied by one another, and the base is the same, the exponents can be added together.

The first step is to try to create a common base.

$4^{x}\cdot2^{5}$

Given that the square of 2 is for, the expression can be rewritten as:

$2^{2}^{x}\cdot 2^{5}$

$2^{2x+5}$

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Question

What is the value of the expression:

$x^{3}\cdotx^{y}\cdot x^{2+y}$

Answer

When values, having the same base, are multiplied by one another, the exponents are added together and the base stays the same.

Thus,

$x^{3}\cdotx^{y}\cdot x^{2+y}$

is equal to

$x^{4}\cdotx^{y+2}$

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Question

Simplify: $x^{7}\cdot2x^{3}$

Answer

When multiplying exponents, the exponents are added together. Thus, 3 and 7 are added together for a sum of 10. In this problem, the "2" becomes a coefficient in front of the x. Therefore, the correct answer is:

$2x^{10}$

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Question

Simplify the expression:

$3x^{4}+6x^{3}-9x^{2}$

Answer

To simplify this problem we need to factor out a

$3x^{4}+6x^{3}-9x^{2}$

$=3x^{2}(x^{2}+2x-3)$

We can do this because multiplying exponents is the same as adding them. Therefore,

$=3x^{2}(x^{2}+2x-3)$

$=3x^2 \cdot x^2 +3x^2\cdot2x -3x^2\cdot3=3x^{2+2}+3\cdot2x^{2+1}-3\cdot 3x^2$

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Question

Simplify:

$x^{6}\cdot x^{4}$

Answer

When multiplying exponents, the exponents are added together.

$x^{6}\cdot x^{4} = x^{6+4}=x^{10}$

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Question

Simplify:

$2x^{2}\cdot 3x^{4}$

Answer

When multiplying exponents, the exponents are added together.

$2x^{2}\cdot 3x^{4}=6x^{2+4}=6x^{6}$

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Question

Solve: $3^5 \cdot 3^{-8}$

Answer

It is not necessary to evaluate both terms and multiply.

According to the rules of exponents, when we have the same bases raised to some power that are multiplied with each other, we can add the powers.

$3^5 \cdot 3^{-8} = 3^{5+( -8)} = 3^{-3}$

This term can be rewritten as a fraction.

$3^{-3 } = \frac{1}{3^3} = \frac{1}{27}$

The answer is: $\frac{1}{27}$

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