How to multiply exponential variables

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ISEE Upper Level Quantitative Reasoning › How to multiply exponential variables

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Fill in the box to form a perfect square trinomial:

$x ^{2} - 13x + \square$

$\frac{169}{4}$

CORRECT

$\frac{169}{2}$

Explanation

To obtain the constant term of a perfect square trinomial, divide the linear coefficient, which here is , by 2, and square the quotient. The result is

$\left (\frac{ -13}{2} \right) ^{2} = \frac{(-13)^{2}}{2^{2}}= \frac{169}{4}$

Factor completely:

$3x^{2} - x -14$

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

CORRECT

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

$\left ( x-2 \right ) \left (3x- 7 \right )$

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

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

Explanation

A trinomial whose leading term has a coefficent other than 1 can be factored using the -method. We split the middle term using two numbers whose product is and whose sum is . These numbers are , so:

$3x^{2} - x -14$

$= 3x^{2} - 7x +6x -14$

$= \left (3x^{2} - 7x \right ) +\left ( 6x -14 \right )$

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

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

Simplify:

$\left (x + \frac{1}{3} \right )^{3}$

$x ^{3 } + x^{2} + \frac{1}{3} x +\frac{1}{27}$

CORRECT

$x ^{3 } + \frac{1}{3} x^{2} + \frac{1}{9} x +\frac{1}{27}$

$x ^{3 } + \frac{2}{3} x^{2} + \frac{2}{9} x +\frac{1}{27}$

$x ^{3 } + 2 x^{2} + \frac{2}{3} x +\frac{1}{27}$

$x ^{3 } + 3 x^{2} +x +\frac{1}{27}$

Explanation

The cube of a sum pattern can be applied here:

$\left (x + \frac{1}{3} \right )^{3}$

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

$= x ^{3 } + x^{2} +3\cdot x \cdot \frac{1}{9} +\frac{1}{27}$

$= x ^{3 } + x^{2} + \frac{1}{3} x +\frac{1}{27}$

Simplify:

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

$256x^{4} +4y ^{2}$

CORRECT

$16x^{4} +4y ^{2}$

$64x^{4} y ^{2}$

$1,024x^{4} y ^{2}$

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

Explanation

$\left (4x \right )^{4} +\left ( 2y \right )^{2} =4^{4}x^{4} + 2^{2}y^{2}=256x^{4} +4y ^{2}$

Multiply:

$\left ( x -3y + 1\right ) \left ( x - 3y - 1\right )$

$x^{2} - 6xy +9y ^{2} - 1$

CORRECT

$x^{2} +9y ^{2} - 1$

$x^{2} + 6xy -9y ^{2} - 1$

$x^{2} + 3xy -9y ^{2} - 1$

$x^{2} - 3xy +9y ^{2} - 1$

Explanation

This can be achieved by using the pattern of difference of squares:

$\left ( x -3y + 1\right ) \left ( x - 3y - 1\right )$

$=\left [ \left ( x -3y \right ) + 1\right ] \left \left[( x - 3y \right ) - 1\right ]$

$= ( x -3y ) ^{2} - 1 ^{2}$

$= (x -3y) ^{2} - 1$

Applying the binomial square pattern:

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

$= x^{2} - 6xy +9y ^{2} - 1$

Expand: $(x-5)^{3}$

$x^{3}-15x^{2}+75x-125$

CORRECT

$x^{3}+15x^{2}-75x-125$

$x^{3}-25x^{2}+50x-125$

$x^{3}+25x^{2}-50x-125$

$x^{3}-125$

Explanation

A binomial can be cubed using the pattern:

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

Set

$(x-5)^{3}$

$= x^{3} - 3x^{2} \cdot 5+ 3x\cdot 5^{2} - 5^{3}$

$= x^{3} - 15x^{2} + 75x - 125$

and are positive integers greater than 1.

Which is the greater quantity?

(A) $(x+y -1)^{2}$

(B) $(x-y+1)^{2}$

(A) is greater

CORRECT

(B) is greater

(A) and (B) are equal

It is impossible to determine which is greater from the information given

Explanation

One way to look at this problem is to substitute . Since , must be positive, and this problem is to compare $(x+u)^{2}$ and $(x-u)^{2}$ .

$(x+u)^{2} = x^{2} + 2ux + u^{2}$

and

$(x-u)^{2} = x^{2} - 2ux + u^{2}$

Since 2, , and are positive, by closure, , and by the addition property of inequality,

$x^{2} + 2ux + u^{2} > x^{2} - 2ux + u^{2}$

$(x+u)^{2} > (x-u)^{2}$

Substituting back:

$(x+y -1)^{2} > (x-y+1)^{2}$

(A) is the greater quantity.

Which is the greater quantity?

(a) $y ^{2} + 4y + 4$

(b) $y^{2} + 6y + 9$

It is impossible to tell from the information given.

CORRECT

(a) is greater.

(a) and (b) are equal.

(b) is greater.

Explanation

We show that either polynomial can be greater by giving two cases:

Case 1:

$y ^{2} + 4y + 4 = 0 ^{2} + 4 \cdot 0 + 4 = 4$

$y^{2} + 6y + 9 = 0^{2} + 6 \cdot 0 + 9 = 9$

$y^{2} + 6y + 9 > y ^{2} + 4y + 4$

Case 2:

$y ^{2} + 4y + 4 = (-3) ^{2} + 4 \cdot (-3) + 4 = 9-12 + 4 = 1$

$y^{2} + 6y + 9 = (-3)^{2} + 6 \cdot (-3) + 9 = 9 - 18 + 9 = 0$

$y ^{2} + 4y + 4 < y^{2} + 6y + 9$

Factor completely:

$4 y^{2} - 100 y$

CORRECT

$4y ^{2}(y - 25)$

$4 (y - 5)^{2}$

Explanation

The greatest common factor of the terms in $4 y^{2} - 100 y$ is , so factor that out:

$4 y^{2} - 100 y$

$= 4y\left ( \frac{4 y^{2}}{4y} - \frac{100 y}{4y} \right )$

$= 4y\left (y - 25 \right )$

Since all factors here are linear, this is the complete factorization.

Simplify:

$\frac{4x^{3}}{y} \cdot \frac{y^{5}}{8x} \cdot xy$

$\frac{ x^{3 } y^{5} }{2 }$

CORRECT

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

$\frac{ x^{3 } y^{4} }{2 }$

$\frac{ x^{2 } y^{5} }{2 }$

Explanation

$\frac{4x^{3}}{y} \cdot \frac{y^{5}}{8x} \cdot xy$

$= \frac{4x^{3}}{y} \cdot \frac{y^{5}}{8x} \cdot \frac{xy}{1}$

$= \frac{4x^{3} \cdot y^{5} \cdot xy }{y \cdot 8x \cdot 1 }$

$= \frac{4x^{3+1 } \cdot y^{5+1} }{8xy }$

$= \frac{4x^{4 } \cdot y^{6} }{8xy }$

$= \frac{ x^{4-1 } \cdot y^{6-1} }{2 }$

$= \frac{ x^{3 } y^{5} }{2 }$

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