How to multiply exponential variables - ISEE Upper Level Quantitative Reasoning

Card 0 of 188

Question

Simplify:

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

Answer

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

Compare your answer with the correct one above

Question

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

Answer

A binomial can be cubed using the pattern:

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

Set

$(x-5)^{3}$

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

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

Compare your answer with the correct one above

Question

Multiply:

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

Answer

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

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

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

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

$= ( x + y ) ^{2} - 9$

Applying the binomial square pattern:

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

Compare your answer with the correct one above

Question

Factor completely:

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

Answer

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:

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

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

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

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

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

Compare your answer with the correct one above

Question

Fill in the box to form a perfect square trinomial:

$x^{2} - 18x + \square$

Answer

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

$\left (\frac{-18}{2} \right )^{2} =\left ( -9\right ) ^{2} = 81$

Compare your answer with the correct one above

Question

Simplify:

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

Answer

The cube of a sum pattern can be applied here:

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

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

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

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

Compare your answer with the correct one above

Question

Fill in the box to form a perfect square trinomial:

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

Answer

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

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

Compare your answer with the correct one above

Question

Expand: $(x + 2)^{10}$

Which is the greater quantity?

(a) The coefficient of $x^{4}$

(b) The coefficient of $x^{6}$

Answer

Using the Binomial Theorem, if $(x + 2)^{n}$ is expanded, the $x^{r}$ term is

$_{n}\textrm{C} _{r} \cdot x^{r} \cdot 2 ^{n-r}$

$= \frac{n!}{(n-r)!r!} \cdot 2 ^{n-r} \cdot x^{r}$ .

This makes $\frac{n!}{(n-r)!r!} \cdot 2 ^{n-r}$ the coefficient of $x^{r}$ .

We compare the values of this expression at for both and :

(a) $\frac{10!}{(10-4)!4!} \cdot 2 ^{10-4} = \frac{10!}{6!4!} \cdot 2 ^{6}$

(b) $\frac{10!}{(10-6)!6!} \cdot 2 ^{10-6} = \frac{10!}{4!6!} \cdot 2 ^{4} = \frac{10!}{6!4!} \cdot 2 ^{4}$

(a) is the greater quantity.

Compare your answer with the correct one above

Question

Which is the greater quantity?

(a) $16x^{2} - 72x + 81$

(b) 8

Answer

$16x^{2} - 72x + 81$

$= 4^{2} x^{2} -2\cdot 4x \cdot 9 + 9^{2}$

$=\left ( 4 x \right ) ^{2} - 2 \cdot 4x \cdot 9 + 9^{2}$

$= \left ( 4x -9 \right )^{2}$

Since ,

$4x - 9 > 4 \cdot3 - 9 = 3$ , so

$16x^{2} - 72x + 81= \left ( 4x -9 \right )^{2} > 3^{2} = 9 >8$

making (a) greater.

Compare your answer with the correct one above

Question

Which is the greater quantity?

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

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

Answer

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:

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

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

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

Compare your answer with the correct one above

Question

and are positive integers. Which is the greater quantity?

(A) $A^{2}+B^{2}$

(B) $(A+B)^{2}$

Answer

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

Since and are positive,

$2AB +A^{2}+B^{2}> 0 +A^{2}+B^{2}$

$A^{2}+2AB +B^{2}> A^{2}+B^{2}$

$(A+B)^{2} > A^{2}+B^{2}$ for all positive and , making (B) greater.

Compare your answer with the correct one above

Question

and are positive integers. Which is the greater quantity?

(A) $A^{2}-B^{2}$

(B) $(A-B)^{2}$

Answer

It is impossible to tell which is greater.

Case 1:

Then

$A^{2}-B^{2} =4^{2}-4^{2} = 16-16 = 0$

and

$(A-B)^{2} = (4-4)^{2} = 0^{2} = 0$ .

This makes (A) and (B) equal.

Case 2:

Then

$A^{2}-B^{2} =5^{2}-4^{2} = 25-16 = 9$

and

$(A-B)^{2} = (5-4)^{2} = 1^{2} = 1$ .

This makes (A) the greater quantity.

Compare your answer with the correct one above

Question

and are negative integers. Which is the greater quantity?

(A) $A^{2}+B^{2}$

(B) $(A+B)^{2}$

Answer

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

Since and are both negative,

.

$2AB +A^{2}+B^{2} > 0 +A^{2}+B^{2}$

$A^{2}+2AB +B^{2} > A^{2}+B^{2}$

$(A+B)^{2} > A^{2}+B^{2}$ for all negative and , making (B) greater.

Compare your answer with the correct one above

Question

and are positive integers greater than 1.

Which is the greater quantity?

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

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

Answer

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.

Compare your answer with the correct one above

Question

and are positive integers greater than 1.

Which is the greater quantity?

(A) $\left (x + y + 1\right ) ^{2}$

(B)

Answer

One way to look at this problem is to substitute . The expressions to be compared are

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

and

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

Since is positive, so is $y^{2}$ , and

$u^{2}> u^{2}- y^{2}$

Substituting back,

$\left (x + y + 1\right ) ^{2} > (x+ 1 ) (x + 2y + 1)$ ,

making (A) greater.

Compare your answer with the correct one above

Question

and are positive integers greater than 1.

Which is the greater quantity?

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

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

Answer

Case 1:

Then

$\left (x + y + 4\right ) ^{2} = \left (2 + 2 + 4\right ) ^{2} = 8 ^{2} = 64$

and

$\left (x + 2y + 2\right ) ^{2} = \left (2 + 2 \cdot 2 + 2\right ) ^{2} = 8 ^{2} = 64$

This makes the quantities equal.

Case 2:

Then

$\left (x + y + 4\right ) ^{2} = \left (2 + 3 + 4\right ) ^{2} = 9 ^{2} = 81$

and

$\left (x + 2y + 2\right ) ^{2} = \left (2 + 2 \cdot 3 + 2\right ) ^{2} = 10 ^{2} = 100$

This makes (B) greater.

Therefore, it is not clear which quantity, if either, is greater.

Compare your answer with the correct one above

Question

Factor:

$4x ^{2} - 9 y^{2} + 14x - 21y$

Answer

We can rewrite as follows:

$4x ^{2} - 9 y^{2} + 14x - 21y$

$= (4x ^{2} - 9 y^{2}) +( 14x - 21y)$

Each group can be factored - the first as the difference of squares, the second as a pair with a greatest common factor. This becomes

,

which, by distribution, becomes

Compare your answer with the correct one above

Question

is a positive number; is the additive inverse of .

Which is the greater quantity?

(a) $a ^{2} - b^{2}$

(b)

Answer

If is the additive inverse of , then, by definition,

.

$a^{2} - b^{2}$ , as the difference of the squares of two expressions, can be factored as follows:

$a^{2} - b^{2} = (a+b) (a-b )$

Since , it follows that

$a^{2} - b^{2} = 0 \cdot (a-b ) = 0$

Another consequence of being the additive inverse of is that

, so

is positive, so is as well.

It follows that $a - b > a ^{2} - b^{2}$ .

Compare your answer with the correct one above

Question

Write in expanded form:

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

Answer

The cube of a sum pattern can be applied here:

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

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

$= 2^{3}\cdot x^{3} + 3\cdot 2^{2}\cdot \frac{1}{2} \cdot x^{2} + 3\cdot2 \cdot \left (\frac{1}{2} \right ) ^{2} \cdot x \left + \left ( \frac{1}{2} \right ) ^{3}$

$= 8x^{3} + 6 x^{2} + \frac{3}{2} x + \frac{1}{8}$

Compare your answer with the correct one above

Question

Simplify the following:

$z^{2}\times z^{5}$

Answer

To multiply variables with exponents, add the exponents. So,

$z^{2}\times z^{5}=z^{7}$

A longer way would be to write out all the multiplies the exponent tells us to do. This is a little clearer on why adding the exponents works but takes longer and isn't necessary once you understand the process.

$z^{2}\times z^{5}= zzzzzzz= z^{7}$

Compare your answer with the correct one above

Tap the card to reveal the answer