SAT Polynomials

To successfully work with polynomials on the SAT, you need to know:

• the difference of squares identity: $a^{2}-b^{2}=(a+b)(a-b)$
• the perfect square identity: $(a \pm b)^{2}=a^{2} \pm 2ab + b^{2}$
• how to find common factors
• how to factor quadratics

If $(x+y)^{2} = 10$ and $(x-y)^{2}=6$, what is $4xy$?

(A) $\ \ -4$
(B) $\ \ 1$
(C) $\ \ 4$
(D) $\ \ 10$
(E) $\ \ 16$

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Correct Answer: C

Solution 1:

Tip: $(a \pm b)^{2} = a^{2} \pm 2ab + b^{2}$
We apply the perfect square formula to each equation.

$(x+y)^{2} = 10 \implies x^{2}+2xy+y^{2}= 10 \quad (1)$

$(x-y)^{2} = 6 \implies \ \ x^{2}-2xy+y^{2 }= 6\ \ \quad (2)$

Subtracting $(2)$ from $(1)$, we get $0+ 4xy + 0=4$. Therefore, $4xy=4.$

Solution 2:

Tip: $a^{2}-b^{2}=(a-b)(a+b)$
This solution is intuitive only if you are very familiar with the difference of squares identity and the perfect squares identity.

We notice that $(x+y)^{2}$ and $(x-y)^{2}$ are perfect squares. Therefore, we can apply the difference of squares formula to get:

$$\begin{array}{lcl} (x+y)^{2}-(x-y)^{2}&=&10-6\\ [x+y+(x-y)][(x+y-(x-y)]&=&4\\ (x+y+x-y)(x+y-x+y)&=&4\\ 2x \cdot 2y&=&4\\ 4xy&=&4\\ \end{array}$$

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Incorrect Choices:

(A)
Refer to Solution 1 above. If in subtracting $(2)$ from $(1)$, you mistakenly get this $0+ 4xy + 0=-4$, instead of this $0+ 4xy + 0=4$, you will get this wrong answer. Similarly, if in Solution 2 you begin with this $(x+y)^{2}-(x-y)^{2}=6-10$ instead of this $(x+y)^{2}-(x-y)^{2}=10-6$, you will get this wrong answer.

(B)
Tip: Read the entire question carefully.
If you are solving for $xy$ instead of $4xy$ you will get this wrong answer.

(D)
Tip: Just because a number appears in the question doesn't mean it is the answer.
This choice is offered to confuse you.

(E)
If you are using the method shown in Solution 1, and if you accidentally add the right hand sides of the equations $(1)$ and $(2)$, instead of subtract them, you will get this wrong answer.

If $a^{2}b^{2}=3$ and $ac-bd=4$, then which of the following equals $a^{3}b^{2}c-a^{2}b^{3}d$?

(A) $\ \ 1$
(B) $\ \ 3$
(C) $\ \ 7$
(D) $\ \ 12$
(E) $\ \ 24$

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Correct Answer: D

Solution:

We begin with $a^{3}b^{2}c-a^{2}b^{3}d$. The greatest common factor of the two terms is $a^{2}b^{2}$. We factor it out:

$a^{3}b^{2}c-a^{2}b^{3}d = a^{2}b^{2}(ac-bd).$

We substitute the given $a^{2}b^{2}=3$ and $ac-bd=4$ and obtain the answer:

$a^{3}b^{2}c-a^{2}b^{3}d = a^{2}b^{2}(ac-bd) = 3 \cdot 4=12.$

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Incorrect Choices:

(A), (B), and (C)
Tip: Just because a number appears in the question doesn't mean it is the answer.
If you don't know how to approach the problem, you might be tempted to choose this answer because it seems familiar -- (A) is the difference of the two numbers that appear in the prompt, (B) is a number that appears in the prompt, and (C) is the sum of the two numbers that appear in the prompt.

(E)
Intended to make you doubt your answer, this choice is double the actual answer.

Which of the following is equivalent to $(x-3)(x+5)-(x-7)(x-6)?$

(A) $\ \ -11x+27$
(B) $\ \ x-57$
(C) $\ \ 15x+57$
(D) $\ \ 15x-57$
(E) $\ \ 2x^{2}-11x+27$

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Correct Answer: D

Solution 1:

Tip: Be careful with signs!
To multiply each pair of binomials, we use a technique, which you may know by the mnemonic FOIL:

First = Multiply the first terms of each binomial
Outside = Multiply the outside terms of each binomial
Inside = Multiply the inside terms of each binomial
Last = Multiply the last terms of each binomial

We demonstrate it with each product in the prompt:

$$\begin{array}{lcr} (x-3)(x+5) &=& \underbrace{x^{2}}_{\text{First}}\ \ \underbrace{+5x}_{\text{Outside}} \ \ \underbrace{-3x}_{\text{Inside}}\ \ \underbrace{-15}_{\text{Last}} \end{array}$$

$$\begin{array}{lcr} (x-7)(x-6) &=& \underbrace{x^{2}}_{\text{First}}\ \ \underbrace{-6x}_{\text{Outside}} \ \ \underbrace{-7x}_{\text{Inside}}\ \ \underbrace{+42}_{\text{Last}} \end{array}$$

Now, to solve the problem, we FOIL:

$$\begin{array}{lcr} &(x-3)(x+5)-(x-7)(x-6) = \\ &=x^{2}+5x-3x-15-[x^{2}-6x-7x+42] &(1)\\ &=x^{2}+2x-15-[x^{2}-13x+42] &(2)\\ &=x^{2}+2x-15-x^{2}+13x-42& (3)\\ &=15x-57 &(4) \end{array}$$

Solution 2:

Tip: When distributing, be careful with signs!
Use the distributive property to expand each product:

$$\begin{array}{lcr} &(x-3)(x+5)-(x-7)(x-6) = \\ &=(x-3)x+5(x-3) -(x-7)x+6(x-7)&(1)\\ &=x^2-3x+5x-15-x^2+7x+6x-42 &(2)\\ &=15x-57& (3)\\ \end{array}$$

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Incorrect Choices:

(A)
Tip: When distributing, be careful with signs!
Refer to Solution 1 above. You will get this wrong answer if in step $(1)$ you omit the brackets, forgetting to distribute the negative sign through all the terms, like this:

$$\begin{array}{lcr} &(x-3)(x+5)-(x-7)(x-6)\\ &=x^{2}+5x-3x-15-x^{2}\fbox{-}6x\fbox{-}7x\fbox{+}42 &(1) \end{array}$$

(B)
Tip: When distributing, be careful with signs!
Refer to Solution 2 above. You will get this wrong answer if in step $(2)$ you don't distribute the negative sign correctly, like this:

$$\begin{array}{lcr} &(x-3)x+5(x-3) \fbox{-}(x-7)x+6(x-7)&(1)\\ &=x^2-3x+5x-15-x^2\fbox{-}7x+6x-42 &(2) \end{array}$$

(D)
Tip: Select the answer with the correct sign!

(E)
Tip: Read the entire question carefully.
Refer to Solution 1 above. If in step $(1)$ you add instead of subtract, like this:

$$\begin{aligned} (x-3)(x+5)\fbox{+}(x-7)(x-6), \end{aligned}$$

you will get this wrong answer.

If you thought these examples difficult and you need to review the material, these links will help:

• Combining Like Terms
• Distributive Property
  $a(b+c)=ab+ac$
• Distributive Property Multiple Terms
• Multiplying Polynomials
• Quadratics-Factoring-Easy
• Factoring Polynomials
• Applying the Difference of Two Squares Identity
  $a^{2}-b^{2}=(a+b)(a-b)$
• Applying the Perfect Square Identity
  $(a \pm b)^{2}=a^{2} \pm 2ab + b^{2}$
• Rules of Exponents

• $a^{2}-b^{2}=(a-b)(a+b)$
• $(a \pm b)^{2} = a^{2} \pm 2ab + b^{2}$
• Be careful with signs!
• SAT General Tips