Polynomials - SAT Math

Card 0 of 264

Question

Subtract from .

Answer

Step 1: We need to read the question carefully. It says subtract from. When you see the word "from", you read the question right to left.

I am subtracting the left equation from the right equation.

Step 2: We need to write the equation on the right minus the equation of the left.

Step 3: Distribute the minus sign in front of the parentheses:

Step 4: Combine like terms:

Step 5: Put all the terms together, starting with highest degree. The degree of the terms is the exponent. Here, the highest degree is 2 and lowest is zero.

The final equation is

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Question

Define an operation $\heartsuit$ on the set of real numbers as follows:

For all real ,

$m \heartsuit n = \left ( m-4n\right )^{3}$

How else could this operation be defined?

Answer

$\left ( m-4n\right )^{3}$ , as the cube of a binomial, can be rewritten using the following pattern:

$\left ( m-4n\right )^{3} = m ^{3} - 3 \cdot m ^{2} \cdot 4n + 3 \cdot m \cdot (4n) ^{2}- (4n) ^{3}$

Applying the rules of exponents to simplify this:

$\left ( m-4n\right )^{3} = m ^{3} - 3 \cdot 4 \cdot m ^{2} \cdot n + 3 \cdot m \cdot 4 ^{2}\cdot n ^{2}- 4^{3} \cdot n ^{3}$

$\left ( m-4n\right )^{3} = m ^{3} - 3 \cdot 4 \cdot m ^{2} \cdot n + 3 \cdot 16 \cdot m\cdot n ^{2}- 4^{3} \cdot n ^{3}$

$\left ( m-4n\right )^{3} = m ^{3} -12 m ^{2} n + 48mn ^{2}- 64n ^{3}$

Therefore, the correct choice is that, alternatively stated,

$m \heartsuit n = m ^{3} -12 m ^{2} n + 48mn ^{2}- 64n ^{3}$ .

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Question

If 〖(x+y)〗2 = 144 and 〖(x-y)〗2 = 64, what is the value of xy?

Answer

We first expand each binomial to get x2 + 2xy + y2 = 144 and x2 - 2xy + y2 = 64. We then subtract the second equation from the first to find 4xy = 80. Finally, we divide each side by 4 to find xy = 20.

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Question

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

and

$G(x) = x^{2} + 5$

What is ?

Answer

$(FG)(x) = F(x)G(x)$ so we multiply the two function to get the answer. We use $x^{m}x^{n} = x^{m+n}$

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Question

Find the product:

Answer

Find the product:

Step 1: Use the distributive property.

Step 2: Combine like terms.

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Question

and represent positive quantities.

$x^{2} = 20$

$y^{2} = 5$

Evaluate $(x-y) (x^{2}+xy+y^{2})$ .

Answer

$(x-y) (x^{2}+xy+y^{2})$ can be recognized as the pattern conforming to that of the difference of two perfect cubes:

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

Additionally,

$x^{2} = 20$ and is positive, so

$x = \sqrt{20}$

Using the product of radicals property, we see that

$x = \sqrt{4 \cdot 5} = \sqrt{4 } \cdot \sqrt{ 5} = 2 \sqrt{ 5}$

and

$x ^{3 } = x^{2} \cdot x = 20 \cdot 2 \sqrt{5} = 40 \sqrt{5}$

$y^{2} = 5$ and is positive, so

$y = \sqrt{5}$ ,

and

$y^{3}= y^{2} \cdot y = 5 \sqrt{5}$

Substituting for $x ^{3 }$ and $y^{3}$ , then collecting the like radicals,

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

$= x^{3} - y^{3}$

$= 40 \sqrt{5}-5 \sqrt{5}$

$=( 40 -5 )\sqrt{5}$

$=35 \sqrt{5}$ .

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Question

represents a positive quantity; represents a negative quantity.

$x^{4} = 116$

$y^{2} = 11$

Evaluate $(x- y)(x+y)(x^{2} + y^{2})$

Answer

The first two binomials are the difference and the sum of the same two expressions, which, when multiplied, yield the difference of their squares:

$(x- y)(x+y)(x^{2} + y^{2})$

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

Again, a sum is multiplied by a difference to yield a difference of squares, which by the Power of a Power Property, is equal to:

$(x^{2}) ^{2} - (y^{2}) ^{2}$

$= x^{2\cdot 2} - y^{2\cdot 2}$

$= x^{4} - y^{4}$

$y^{2} = 11$ , so by the Power of a Power Property,

$(y^{2} )^{2}= 11^{2}$

$y^{2\cdot 2} = 121$

$y^{4} = 121$

Also, $x^{4} = 116$ , so we can now substitute accordingly:

$(x- y)(x+y)(x^{2} + y^{2})$

$= x^{4} - y^{4}$

Note that the signs of and are actually irrelevant to the problem.

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Question

represents a positive quantity; represents a negative quantity.

$x^{6} = 625$

$y ^{6} = 64$

Evaluate $(x-y) (x^{2}+xy+y^{2})$ .

Answer

$(x-y) (x^{2}+xy+y^{2})$ can be recognized as the pattern conforming to that of the difference of two perfect cubes:

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

Additionally, by way of the Power of a Power Property,

$(x^{3})^{ 2 } = x^{3 \cdot 2} = x^{6 }$ , making $x^{3}$ a square root of $x^{6}$ , or 625; since is positive, so is $x^{3}$ , so

$x^{3} = \sqrt{625} = 25$ .

Similarly, $y ^{3}$ is a square root of $y ^{6}$ , or 64; since is negative, so is $y^{3}$ (as an odd power of a negative number is negative), so

$y^{3} = - \sqrt{64} = -8$ .

Therefore, substituting:

$(x-y) (x^{2}+xy+y^{2}) = x^{3} - y^{3} = 25 - (-8) = 25+8= 33$ .

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Question

If 〖(x+y)〗2 = 144 and 〖(x-y)〗2 = 64, what is the value of xy?

Answer

We first expand each binomial to get x2 + 2xy + y2 = 144 and x2 - 2xy + y2 = 64. We then subtract the second equation from the first to find 4xy = 80. Finally, we divide each side by 4 to find xy = 20.

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Question

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

and

$G(x) = x^{2} + 5$

What is ?

Answer

$(FG)(x) = F(x)G(x)$ so we multiply the two function to get the answer. We use $x^{m}x^{n} = x^{m+n}$

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Question

Find the product:

Answer

Find the product:

Step 1: Use the distributive property.

Step 2: Combine like terms.

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Question

and represent positive quantities.

$x^{2} = 20$

$y^{2} = 5$

Evaluate $(x-y) (x^{2}+xy+y^{2})$ .

Answer

$(x-y) (x^{2}+xy+y^{2})$ can be recognized as the pattern conforming to that of the difference of two perfect cubes:

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

Additionally,

$x^{2} = 20$ and is positive, so

$x = \sqrt{20}$

Using the product of radicals property, we see that

$x = \sqrt{4 \cdot 5} = \sqrt{4 } \cdot \sqrt{ 5} = 2 \sqrt{ 5}$

and

$x ^{3 } = x^{2} \cdot x = 20 \cdot 2 \sqrt{5} = 40 \sqrt{5}$

$y^{2} = 5$ and is positive, so

$y = \sqrt{5}$ ,

and

$y^{3}= y^{2} \cdot y = 5 \sqrt{5}$

Substituting for $x ^{3 }$ and $y^{3}$ , then collecting the like radicals,

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

$= x^{3} - y^{3}$

$= 40 \sqrt{5}-5 \sqrt{5}$

$=( 40 -5 )\sqrt{5}$

$=35 \sqrt{5}$ .

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Question

represents a positive quantity; represents a negative quantity.

$x^{4} = 116$

$y^{2} = 11$

Evaluate $(x- y)(x+y)(x^{2} + y^{2})$

Answer

The first two binomials are the difference and the sum of the same two expressions, which, when multiplied, yield the difference of their squares:

$(x- y)(x+y)(x^{2} + y^{2})$

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

Again, a sum is multiplied by a difference to yield a difference of squares, which by the Power of a Power Property, is equal to:

$(x^{2}) ^{2} - (y^{2}) ^{2}$

$= x^{2\cdot 2} - y^{2\cdot 2}$

$= x^{4} - y^{4}$

$y^{2} = 11$ , so by the Power of a Power Property,

$(y^{2} )^{2}= 11^{2}$

$y^{2\cdot 2} = 121$

$y^{4} = 121$

Also, $x^{4} = 116$ , so we can now substitute accordingly:

$(x- y)(x+y)(x^{2} + y^{2})$

$= x^{4} - y^{4}$

Note that the signs of and are actually irrelevant to the problem.

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Question

represents a positive quantity; represents a negative quantity.

$x^{6} = 625$

$y ^{6} = 64$

Evaluate $(x-y) (x^{2}+xy+y^{2})$ .

Answer

$(x-y) (x^{2}+xy+y^{2})$ can be recognized as the pattern conforming to that of the difference of two perfect cubes:

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

Additionally, by way of the Power of a Power Property,

$(x^{3})^{ 2 } = x^{3 \cdot 2} = x^{6 }$ , making $x^{3}$ a square root of $x^{6}$ , or 625; since is positive, so is $x^{3}$ , so

$x^{3} = \sqrt{625} = 25$ .

Similarly, $y ^{3}$ is a square root of $y ^{6}$ , or 64; since is negative, so is $y^{3}$ (as an odd power of a negative number is negative), so

$y^{3} = - \sqrt{64} = -8$ .

Therefore, substituting:

$(x-y) (x^{2}+xy+y^{2}) = x^{3} - y^{3} = 25 - (-8) = 25+8= 33$ .

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Question

Solve for .

$\frac{x^{2}+2x-8}{x^{2}+5x-14}=\frac{3}{4}$

Answer

$\frac{x^{2}+2x-8}{x^{2}+5x-14}=\frac{3}{4}$

Factor the expression

numerator: find two numbers that add to 2 and multiply to -8 \[use 4,-2\]

denominator: find two numbers that add to 5 and multiply to -14 \[use 7,-2\]

new expression:

$\frac{(x+4)(x-2)}{(x+7)(x-2)}=\frac{3}{4}$

Cancel the and cross multiply.

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Question

Simplify the following expression:

$(2q^{2}+4q+7)-(3q^{2}-5)$

Answer

This is not a FOIL problem, as we are adding rather than multiplying the terms in parentheses.

Add like terms together:

$2q^{2}-3q^{2}=-1q^{2}=-q^2$

has no like terms.

Combine these terms into one expression to find the answer:

$-q^{2}+4q+12$

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Question

Solve each problem and decide which is the best of the choices given.

What are the zeros of the following trinomial?

Answer

First factor out a . Then the factors of the remaining polynomial,

, are and .

Set everything equal to zero and you get , , and because you cant forget to set equal to zero.

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Question

What is a possible value for x in x2 – 12x + 36 = 0 ?

Answer

You need to factor to find the possible values for x. You need to fill in the blanks with two numbers with a sum of -12 and a product of 36. In both sets of parenthesis, you know you will be subtracting since a negative times a negative is a positive and a negative plus a negative is a negative

(x –)(x –).

You should realize that 6 fits into both blanks.

You must now set each set of parenthesis equal to 0.

x – 6 = 0; x – 6 = 0

Solve both equations: x = 6

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Question

If r and t are constants and x2 +rx +6=(x+2)(x+t), what is the value of r?

Answer

We first expand the right hand side as x2+2x+tx+2t and factor out the x terms to get x2+(2+t)x+2t. Next we set this equal to the original left hand side to get x2+rx +6=x2+(2+t)x+2t, and then we subtract x2 from each side to get rx +6=(2+t)x+2t. Since the coefficients of the x terms on each side must be equal, and the constant terms on each side must be equal, we find that r=2+t and 6=2t, so t is equal to 3 and r is equal to 5.

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Question

2x + 3y = 5a + 2b (1)

3x + 2y = 4a – b (2)

Express x2 – y2 in terms of a and b

Answer

Add the two equations together to yield 5x + 5y = 9a + b, then factor out 5 to get 5(x + y) = 9a + b; divide both sides by 5 to get x + y = (9a + b)/5; subtract the two equations to get x - y = -a - 3b. So, x2 – y2 = (x + y)(x – y) = (9a + b)/5 (–a – 3b) = (–\[(9a)\]2 – 28ab – \[(3b)\]2)/5

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