Binomials - PSAT Math

Card 0 of 84

Solve for .

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

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.

Multiply the binomial.

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By multiplying with the foil method, we multiply our first values giving , our outside values giving . our inside values which gives , and out last values giving .

Give the coefficient of in the binomial expansion of $\left ( 0.2x+ 5 \right $)^{7}$$ .

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If the expression $\left ( A x + B\right $)^{n}$$ is expanded, then by the binomial theorem, the term is

$C(n, k) \cdot\left ( Ax \right $)^{k}$ \cdot B ^{n - k}$

$= C(n, k) \cdot A ^{k} \cdot B ^{n - k} \cdot $x^{k}$$

or, equivalently, the coefficient of is

$C(n, k) \cdot A ^{k} \cdot B ^{n - k}$

Therefore, the coefficient can be determined by setting

$C(7,5) \cdot $0.2^{5}$ \cdot 5 ^{7 - 5}$

$= C(7,5) \cdot 0.2 ^{5} \cdot 5 ^{2}$

$= 21\cdot 0.00032 \cdot 25$

Give the coefficient of in the binomial expansion of $\left (6x+ $\frac{1}{6}\right $)^{7}$$ .

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If the expression $\left ( A x + B\right $)^{n}$$ is expanded, then by the binomial theorem, the term is

$C(n, k) \cdot\left ( Ax \right $)^{k}$ \cdot B ^{n - k}$

$= C(n, k) \cdot A ^{k} \cdot B ^{n - k} \cdot $x^{k}$$

or, equivalently, the coefficient of is

$C(n, k) \cdot A ^{k} \cdot B ^{n - k}$

Therefore, the coefficient can be determined by setting

$A = 6, B = $\frac{1}{6}$, n = 7, k= 4$ :

$C(7,4) \cdot 6 ^{4} \cdot\left ( $\frac{1}{6}$ \right ) ^{7-4}$

$=C(7,4) \cdot 6 ^{4} \cdot\left ( $\frac{1}{6}$ \right ) ^{3}$

$=35 \cdot 1,296 \cdot $\frac{1}{216}$$

Give the coefficient of in the product

$\left ( x+ 0.4\right ) (x - 0.2) (3x-0.7)$ .

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While this problem can be answered by multiplying the three binomials, it is not necessary. There are three ways to multiply one term from each binomial such that two terms and one constant are multiplied; find the three products and add them, as follows:

$\left ($\underline{ x}$+ 0.4\right ) ($\underline{x}$ - 0.2) (3x$\underline{-0.7}$)$

$x \cdot x \cdot (-0.7) = $-0.7x^{2}$$

$\left ($\underline{ x}$+ 0.4\right ) (x $\underline{-0.2}$)($\underline{3x}$-0.7)$

$x \cdot (-0.2) \cdot 3x= $-0.6x^{2}$$

$\left (x+ $\underline{0.4}\right) ($\underline{x}$ - 0.2)($\underline{3x}$-0.7)$

$0.4 \cdot x \cdot 3x = 1.2 $x^{2}$$

Add: .

The correct response is .

Give the coefficient of in the product

$\left (3x- 7 \right ) \left ( 4x+3\right )\left ( 2x-7\right )$ .

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While this problem can be answered by multiplying the three binomials, it is not necessary. There are three ways to multiply one term from each binomial such that two terms and one constant are multiplied; find the three products and add them, as follows:

$\left ($\underline{3x}$- 7 \right ) \left ( $\underline{4x}$+3\right )\left ( 2x$\underline{-7}$\right )$

$3x \cdot 4x \cdot (-7) = $-84x^{2}$$

$\left ($\underline{3x}$- 7 \right ) \left ( 4x$\underline{+3}$\right )\left ( $\underline{2x}$-7\right )$

$3x \cdot 3 \cdot 2x = $18x^{2}$$

$\left (3x$\underline{- 7}$ \right ) \left ( $\underline{4x}$+3\right )\left ( $\underline{2x}$-7\right )$

$-7 \cdot 4x \cdot 2x = $-56x^{2}$$

Add:

The correct response is -122.

Give the coefficient of in the product

$\left ( 2x + $\frac{1}{3}\right) \left( x- $\frac{1}{6}\right ) \left ( x+ $\frac{1}{4}\right )$

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While this problem can be answered by multiplying the three binomials, it is not necessary. There are three ways to multiply one term from each binomial such that two terms and one constant are multiplied; find the three products and add them, as follows:

$\left ( $\underline{2x }$+ $\frac{1}{3}\right) \left( $\underline{x}$- $\frac{1}{6}\right ) \left ( x$\underline{+ \frac{1}{4}$}\right )$

$2x \cdot x \cdot $\frac{1}{4}$ = $$\frac{1}{2}$x^{2}$$

$\left( $\underline{2x }$+ $\frac{1}{3}\right ) \left ( x $\underline{- \frac{1}{6}$}\right ) \left ( $\underline{x}$+ $\frac{1}{4}\right)$

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

$\left ( 2x$\underline{ + \frac{1}{3}$}\right ) \left ( $\underline{x}$- $\frac{1}{6}\right) \left( $\underline{x}$+ $\frac{1}{4}\right )$

$\frac{1}{3}$ \cdot x \cdot x= $$\frac{1}{3}$x^{2}$

Add:

The correct response is $\frac{1}{2}$ .

Give the coefficient of in the binomial expansion of $\left ( 2x+ 0.5\right $)^{8}$$ .

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If the expression $\left ( A x + B\right $)^{n}$$ is expanded, then by the binomial theorem, the term is

$C(n, k) \cdot\left ( Ax \right $)^{k}$ \cdot B ^{n - k}$

$= C(n, k) \cdot A ^{k} \cdot B ^{n - k} \cdot $x^{k}$$

or, equivalently, the coefficient of is

$C(n, k) \cdot A ^{k} \cdot B ^{n - k}$

Therefore, the coefficient can be determined by setting

:

$C(8,5) \cdot $2^{5}$ \cdot 0.5 ^{8-5}$

$=C(8,5) \cdot $2^{5}$ \cdot 0.5 ^{3}$

$= 56\cdot 32 \cdot 0.125$

Decrease by 40%. Which of the following will this be equal to?

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A number decreased by 40% is equivalent to 100% of the number minus 40% of the number. This is taking 60% of the number, or, equivalently, multiplying it by 0.6.

Therefore, decreased by 40% is 0.6 times this, or

$0.6 \left (18d +30 \right ) = 0.6 \cdot 18d + 0.6 \cdot 30 = 10.8d + 18$

Find the product:

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Find the product:

Use the distributive property:

Write the resulting expression in standard form:

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

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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.

Which of these expressions can be simplified further by collecting like terms?

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A binomial can be simplified further if and only if the two terms have the same combination of variables and the same exponents for each like variable. This is not the case in any of the four binomials given, so none of the expressions can be simplified further.

Solve for .

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

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.

Multiply the binomial.

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By multiplying with the foil method, we multiply our first values giving , our outside values giving . our inside values which gives , and out last values giving .

Give the coefficient of in the binomial expansion of $\left ( 0.2x+ 5 \right $)^{7}$$ .

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If the expression $\left ( A x + B\right $)^{n}$$ is expanded, then by the binomial theorem, the term is

$C(n, k) \cdot\left ( Ax \right $)^{k}$ \cdot B ^{n - k}$

$= C(n, k) \cdot A ^{k} \cdot B ^{n - k} \cdot $x^{k}$$

or, equivalently, the coefficient of is

$C(n, k) \cdot A ^{k} \cdot B ^{n - k}$

Therefore, the coefficient can be determined by setting

$C(7,5) \cdot $0.2^{5}$ \cdot 5 ^{7 - 5}$

$= C(7,5) \cdot 0.2 ^{5} \cdot 5 ^{2}$

$= 21\cdot 0.00032 \cdot 25$

Give the coefficient of in the binomial expansion of $\left (6x+ $\frac{1}{6}\right $)^{7}$$ .

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If the expression $\left ( A x + B\right $)^{n}$$ is expanded, then by the binomial theorem, the term is

$C(n, k) \cdot\left ( Ax \right $)^{k}$ \cdot B ^{n - k}$

$= C(n, k) \cdot A ^{k} \cdot B ^{n - k} \cdot $x^{k}$$

or, equivalently, the coefficient of is

$C(n, k) \cdot A ^{k} \cdot B ^{n - k}$

Therefore, the coefficient can be determined by setting

$A = 6, B = $\frac{1}{6}$, n = 7, k= 4$ :

$C(7,4) \cdot 6 ^{4} \cdot\left ( $\frac{1}{6}$ \right ) ^{7-4}$

$=C(7,4) \cdot 6 ^{4} \cdot\left ( $\frac{1}{6}$ \right ) ^{3}$

$=35 \cdot 1,296 \cdot $\frac{1}{216}$$

Give the coefficient of in the product

$\left ( x+ 0.4\right ) (x - 0.2) (3x-0.7)$ .

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While this problem can be answered by multiplying the three binomials, it is not necessary. There are three ways to multiply one term from each binomial such that two terms and one constant are multiplied; find the three products and add them, as follows:

$\left ($\underline{ x}$+ 0.4\right ) ($\underline{x}$ - 0.2) (3x$\underline{-0.7}$)$

$x \cdot x \cdot (-0.7) = $-0.7x^{2}$$

$\left ($\underline{ x}$+ 0.4\right ) (x $\underline{-0.2}$)($\underline{3x}$-0.7)$

$x \cdot (-0.2) \cdot 3x= $-0.6x^{2}$$

$\left (x+ $\underline{0.4}\right) ($\underline{x}$ - 0.2)($\underline{3x}$-0.7)$

$0.4 \cdot x \cdot 3x = 1.2 $x^{2}$$

Add: .

The correct response is .

Give the coefficient of in the product

$\left (3x- 7 \right ) \left ( 4x+3\right )\left ( 2x-7\right )$ .

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While this problem can be answered by multiplying the three binomials, it is not necessary. There are three ways to multiply one term from each binomial such that two terms and one constant are multiplied; find the three products and add them, as follows:

$\left ($\underline{3x}$- 7 \right ) \left ( $\underline{4x}$+3\right )\left ( 2x$\underline{-7}$\right )$

$3x \cdot 4x \cdot (-7) = $-84x^{2}$$

$\left ($\underline{3x}$- 7 \right ) \left ( 4x$\underline{+3}$\right )\left ( $\underline{2x}$-7\right )$

$3x \cdot 3 \cdot 2x = $18x^{2}$$

$\left (3x$\underline{- 7}$ \right ) \left ( $\underline{4x}$+3\right )\left ( $\underline{2x}$-7\right )$

$-7 \cdot 4x \cdot 2x = $-56x^{2}$$

Add:

The correct response is -122.

Give the coefficient of in the product

$\left ( 2x + $\frac{1}{3}\right) \left( x- $\frac{1}{6}\right ) \left ( x+ $\frac{1}{4}\right )$

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While this problem can be answered by multiplying the three binomials, it is not necessary. There are three ways to multiply one term from each binomial such that two terms and one constant are multiplied; find the three products and add them, as follows:

$\left ( $\underline{2x }$+ $\frac{1}{3}\right) \left( $\underline{x}$- $\frac{1}{6}\right ) \left ( x$\underline{+ \frac{1}{4}$}\right )$

$2x \cdot x \cdot $\frac{1}{4}$ = $$\frac{1}{2}$x^{2}$$

$\left( $\underline{2x }$+ $\frac{1}{3}\right ) \left ( x $\underline{- \frac{1}{6}$}\right ) \left ( $\underline{x}$+ $\frac{1}{4}\right)$

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

$\left ( 2x$\underline{ + \frac{1}{3}$}\right ) \left ( $\underline{x}$- $\frac{1}{6}\right) \left( $\underline{x}$+ $\frac{1}{4}\right )$

$\frac{1}{3}$ \cdot x \cdot x= $$\frac{1}{3}$x^{2}$

Add:

The correct response is $\frac{1}{2}$ .

Give the coefficient of in the binomial expansion of $\left ( 2x+ 0.5\right $)^{8}$$ .

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If the expression $\left ( A x + B\right $)^{n}$$ is expanded, then by the binomial theorem, the term is

$C(n, k) \cdot\left ( Ax \right $)^{k}$ \cdot B ^{n - k}$

$= C(n, k) \cdot A ^{k} \cdot B ^{n - k} \cdot $x^{k}$$

or, equivalently, the coefficient of is

$C(n, k) \cdot A ^{k} \cdot B ^{n - k}$

Therefore, the coefficient can be determined by setting

:

$C(8,5) \cdot $2^{5}$ \cdot 0.5 ^{8-5}$

$=C(8,5) \cdot $2^{5}$ \cdot 0.5 ^{3}$

$= 56\cdot 32 \cdot 0.125$