## MAT 122 Course Prep

This collection of problems is meant to provide practice for a range of skills needed for MAT 122. Read and attempt each problem first; if you aren’t sure how to start a problem, explore the resources on the right to refresh your memory and try again. A drop-down button is found beneath each problem for you to compare both your logic and final answers. Keep track of the skills you aren’t comfortable with, and reacquaint yourself with them so you’re fully prepared for the topics you’ll grapple with soon. Get help in the Math Lab.  Return to the Course Prep page.

Problem 1:  Evaluate the following expressions.  Remember to follow the order of operations.

 Part a. $(-5+(9-5(4))+3) \div (5(-2)-1)$ Click here to show solution. Everything before the division sign is part of the numerator, and everything after it is part of the denominator.  In the numerator, the 4 is located inside 3 sets of parentheses, so we must start there.  It is being multiplied by 5, so we replace 5(4) with 20.  In the denominator, the -2 is inside two sets of parentheses, and it is being multiplied by 5, so we replace 5(-2) with -10. $(-5+(9-20+3)) \div (-10-1)$ Again looking at the numerator, inside the inner parentheses, we can evaluate 9 – 20 + 3 = -8.  In the denominator, we can evaluate -10 – 1 = -11. $(-5+(-8)) \div (-11)$ The numerator can be simplified by evaluating -5 + (-8) = -13.  Since both the numerator and denominator are negative, the result will be positive.  Thus, our final answer is: $\boxed{\dfrac{13}{11}}.$ Need Help ? MathPlanet – Order of Operations Monterey – Order of Operations

Problem 2:  Find the value of y using the given value of x.

 Part a. $y = -x^{2} - 2x - 7 \quad \text{and} \quad x=4$ Click here to show solution. Since x = 4, we can replace every x in the first equation with a 4: $y = -(4)^{2} - 2(4) - 7$ Order of operations tells us that we need to square the 4 and then multiply by -1, and also multiply 2 and 4: $y = -16 - 8 - 7$ Thus, our final answer is:  $\boxed{y = -31}.$ Need Help ? MathPlanet – Evaluating Expressions Part b. $y = -x^{2} - 2x - 7 \quad \text{and} \quad x=-3$ Click here to show solution. Since x = -3, we can replace every x in the first equation with a -3: $y = -(-3)^{2} - 2(-3) - 7$ Order of operations tells us that we need to square the -3 first, which means -3 times -3.  This makes 9 because the negatives cancel.  Also, we need to multiply 2 and -3, which makes -6: $y = -(9) - (-6) - 7$ Now we can make the 9 negative, but the 6 ends up positive because the two minus signs cancel: $y = -9 + 6 - 7$ Thus, our final answer is:  $\boxed{y = -10}.$ Part c. $y = (x + 7)(x + 3)(x - 8) \quad \text{and} \quad x=-2$ Click here to show solution. Since x = -2, we can replace every x in the first equation with a -2: $y = (-2 + 7)(-2 + 3)(-2 - 8)$ Order of operations tells us that we need evaluate the insides of every set of parentheses first, so: $y = (5)(1)(-10)$ Now we can multiply these three numbers in any order, so let’s do 5 times 1 is 5, and then multiplying by -10 gives us -50. Thus, our final answer is:  $\boxed{y = -50}.$ Part d. $y = (x + 7)(x + 3)(x - 8) \quad \text{and} \quad x=-3$ Click here to show solution. Since x = -3, we can replace every x in the first equation with a -3: $y = (-3 + 7)(-3 + 3)(-3 - 8)$ Now we must simplify the expressions inside every set of parentheses: $y = (4)(0)(-11)$ This means 4 times 0 times -11. Remember that any number times zero is equal to zero. $\boxed{y = 0}.$

Problem 3:  Use the following graph to answer the questions below.

 Need Help ? MathPlanet – Coordinates & Ordered Pairs Part a. Does the point (6, 5) fall on the graph of the line ? Click here to show solution. A coordinate point tells you the x-coordinate first, followed by the y-coordinate.  We can see that the point (6, 5) has an x-coordinate of 6.  We can see on the x-axis, between 0 and 5, that this region is divided into 5 equal parts, marked by the light grey lines.  This means that each square cell on the grid is one unit wide.  To find 6 on the x-axis, we just need to go one unit to the right of the 5.  Then, we follow the light grey vertical grid-line up to the red line, and we have found the point on the line with x-coordinate 6.  Finally, we follow the light grey horizontal grid-line from this point to the left, until we reach the y-axis.  This horizontal grid-line intersects the y-axis right at the number 5, so we know that 5 is the y-coordinate that corresponds to the x-coordinate of 6. Thus, we have determined that (6, 5) does in fact fall on the graph of the line.

Problem 4:  Complete the following table.

 Part a. $\begin{array}{c|c} x & 3x-1 \\\hline 2 & 5 \\4 & 11 \\7 & \\ & 38 \end{array}$ Click here to show solution. First, we will try to fill in the blank space next to the 7. The 7 is in the x column, so we can set x = 7 and plug this value in for x in the expression 3x – 1. $3(7)-1$ Simplifying, we find that when x = 7: $3x - 1 = 21 - 1 = \boxed{20}.$ Next, we can try to fill the blank space in the first column next to 38. The 38 is in the 3x – 1 column, so we can set 3x – 1 = 38 and solve this equation to find the value of x. Adding 1 to both sides gives us: $3x = 39$ Our next step will be to divide both sides by 3. Thus, we have found that when 3x – 1 = 38: $x = \boxed{13}.$ Our completed table looks like this: $\begin{array}{c|c} x & 3x-1 \\\hline 2 & 5 \\4 & 11 \\7 & 20 \\13 & 38 \end{array}$ Need Help ? MathPlanet – Table Representation of Functions

Problem 5:  Solve the following equations.

 Part a. $100 = 0.2 x$ Click here to show solution. First, we can replace the decimal 0.2 with its fraction equivalent and simplify: $0.2 = \dfrac{2}{10} = \dfrac{1}{5}$ Now we can see that, to isolate x, we need to multiply both sides by 5: $100 = \dfrac{1}{5} x \qquad \longrightarrow \qquad \boxed{500 = x}.$ Need Help ? MathPlanet – Solving Linear Equations Part b. $44 = 8 - 6x$ Click here to show solution. Our goal is to isolate x.  First, we can subtract 8 from both sides, and on the left side, we will get 44 – 8 = 36: $36 = -6x$ Our last step is to divide both sides by -6, which is going to leave x alone on the right side, and will make: $\dfrac{36}{-6} = x$ Finally, we can recall that 36 is 6 times 6, and we can pull the negative sign out front, because a positive divided by a negative results in a negative answer.  Thus, we get: $\boxed{ -6 = x }.$ Part c. $14 + 2x = 9 - 3x$ Click here to show solution. Notice that x appears on both sides, so we need to combine those like terms.  We can add 3x to both sides, which will give us 5x on the left and no x on the right: $14 + 5x = 9$ Now we can get rid of the 14 by subtracting from both sides, and calculating 9 – 14 = -5: $5x = -5$ Our last step is to divide by 5 on both sides: $\boxed{x = -1}.$ Part d. $-80 = \dfrac{4}{3} x$ Click here to show solution. We could deal with the 3 in the denominator first (multiply both sides by 3) and then deal with the 4 in the numerator (divide both sides by 4), but there is another way. Since the fraction 4/3 is the coefficient of x, we can simply divide both sides by 4/3: $\dfrac{-80}{\frac{4}{3}} = x$ Recall that dividing by a fraction is the same as multiplying by the reciprocal: $-80 \cdot \dfrac{3}{4} = x$ We can write -80 as a fraction, and multiply the numerators and denominators separately: $\dfrac{-80}{1} \cdot \dfrac{3}{4} = x \quad \longrightarrow \quad \dfrac{-240}{4} = x$ We can simplify this to get: $\boxed{-60 = x}.$

Problem 6:  Answer the following questions.

 Captain Tim’s boat caught 360 lobsters. Captain Steve’s boat caught 450 lobsters. Need Help ? MathPlanet – Proportions and Percents Part a. The number of lobsters caught by Captain Tim’s boat is what percent of the number of lobsters caught by Captain Steve’s boat ? Click here to show solution. First, we will determine the ratio: $\dfrac{\text{Tim's Lobsters}}{\text{Steve's Lobsters}} = \dfrac{360}{450}$ Simplifying this fraction, we can divide the numerator and denominator by 10: $\dfrac{360}{450} = \dfrac{36}{45}$ And then we can divide the numerator and denominator by 9: $\dfrac{36}{45} = \dfrac{4}{5}$ Now, to get a percent, we want the denominator to equal 100. So we can multiply the numerator and denominator by 20: $\dfrac{4}{5} = \dfrac{80}{100}$ A percent is just the numerator of a fraction with a denominator of 100. Thus, our final answer is:  $\boxed{80\%}.$ Part b. The number of lobsters caught by Captain Steve’s boat is what percent of the number of lobsters caught by Captain Tim’s boat ? Click here to show solution. First, we will determine the ratio: $\dfrac{\text{Steve's Lobsters}}{\text{Tim's Lobsters}} = \dfrac{450}{360}$ Notice that this fraction is simply the reciprocal of the fraction we found in part a. The fraction from part a simplified to 4/5, so this one must simplify to 5/4. $\dfrac{450}{360} = \dfrac{5}{4}$ Now, to get a percent, we want the denominator to equal 100. So we can multiply the numerator and denominator by 25: $\dfrac{5}{4} = \dfrac{125}{100}$ A percent is just the numerator of a fraction with a denominator of 100. Thus, our final answer is:  $\boxed{125\%}.$

Problem 7:  Rewrite each expression in the form:  xa   (this is called base-exponent form).

 Part a. $\dfrac{1}{x^{2}}$ Click here to show solution. We can move a factor from the denominator to the numerator as long as we change the sign of the exponent: $\dfrac{1}{x^{2}} = \dfrac{x^{-2}}{1}$ If it seems like the 1 in the numerator disappeared, remember that multiplying by one won’t change the value of $x^{-2}$.  Dividing by 1 also won’t change the value of $x^{-2}$, so we can just write: $\dfrac{1}{x^{2}} = \boxed{x^{-2}}$ If we want to show all the steps, we could write: $\dfrac{1}{x^{2}} = \dfrac{1}{1 \cdot x^{2}} = \dfrac{1 \cdot x^{-2}}{1} =\dfrac{x^{-2}}{1} = x^{-2}$ Need Help ? BrownMath – Negative and Fractional Exponents MathPlanet – Exponent Rules Part b. $\sqrt{x}$ Click here to show solution. We can rewrite radical expressions using fractional exponents.  For a square root, we use the fraction 1/2: $\sqrt{x} = \boxed{x^{\frac{1}{2}}}$ If this seems strange, think of it this way:  20 = 1  and  21 = 2  so we would expect that  $2^{\frac{1}{2}}$  would produce a number between 1 and 2, and  $\sqrt{2}$  is between 1 and 2. Part c. $\left(\dfrac{1}{x^{3}}\right)^{4}$ Click here to show solution. We can distribute an exponent over division, which means the 4 becomes the exponent of the 1 in the numerator and the x^3 in the denominator: $\left(\dfrac{1}{x^{3}}\right)^{4} = \dfrac{(1)^{4}}{\left(x^{3}\right)^{4}}$ Now 1 multiplied by itself 4 times is still 1, but in the denominator we have to combine the exponents.  Recall that successive exponents can be combined by multiplication: $\left(\dfrac{1}{x^{3}}\right)^{4} = \dfrac{1}{x^{12}}$ Finally, we can rewrite this by moving the x into the numerator, and changing its exponent from positive to negative: $\left(\dfrac{1}{x^{3}}\right)^{4} = \boxed{x^{-12}}$ We could also have written out 4 copies of the original fraction: $\left(\dfrac{1}{x^{3}}\right)^{4} = \dfrac{1}{x^{3}} \cdot \dfrac{1}{x^{3}} \cdot \dfrac{1}{x^{3}} \cdot \dfrac{1}{x^{3}}$ This could then be expanded as: $\left(\dfrac{1}{x^{3}}\right)^{4} = \dfrac{1}{x\cdot x\cdot x} \cdot \dfrac{1}{x\cdot x\cdot x} \cdot \dfrac{1}{x\cdot x\cdot x} \cdot \dfrac{1}{x\cdot x\cdot x}$ Part d. $\sqrt[3]{x^{2}}$ Click here to show solution. Recall that the 3 is called the index of the root, and indicates that this is not a square root but a cube root.  Following our earlier rule, we can rewrite this root using a 1/3 exponent: $\sqrt[3]{x^{2}} = \left(x^{2}\right)^{\frac{1}{3}}$ Exponent rules allow us to multiply successive exponents, so we calculate 2 times 1/3 to get 2/3 as the exponent: $\sqrt[3]{x^{2}} = \boxed{x^{\frac{2}{3}}}.$

Problem 8:  Solve the following problems by setting up a proportion and solving.

 Part a. If 6 dogs require 42 kilograms of kibble for a week’s stay at a kennel, how much kibble is required for a week’s stay for 8 dogs ? Click here to show solution. Since the ratio between amount of food and number of dogs is assumed to be constant, we can set up a proportion, which means setting two fractions equal to each other. We know that 42 kg of food divided among 6 dogs must be equal to some unknown amount of food divided among 8 dogs: $\dfrac{42\text{ kg}}{6\text{ dogs}} = \dfrac{x\text{ kg}}{8\text{ dogs}}$ Now we can simplify the left side, where 42 divided by 6 equals 7: $7 \frac{\text{kg}}{\text{dog}} = \dfrac{x\text{ kg}}{8\text{ dogs}}$ At 7 kilograms per dog, it would take 8 times 7 equals 56 kilograms to feed 8 dogs. Thus, to feed 8 dogs for a week, it would take $\boxed{56 \text{ kg}}$ of kibble. Need Help ? MathPlanet – Proportions and Percents

Problem 9:  Solve the following problem.

 Part a. You ate 3/4 of a frozen lasagna. If the label on the package states that 1/3 of a lasagna is a serving, how many servings of lasagna did you eat ? Click here to show solution 1. If 1/3 of a lasagna is equal to 1 serving, then there are 3 servings in a whole lasagna. We want to know how many servings there are in 3/4 of a lasagna. This is like asking how many times 1/3 goes into 3/4. It may help to write 1/3 and 3/4 with a common denominator, so rewrite 1/3 as 4/12 and rewrite 3/4 as 9/12. Now we can see that 1/3 goes into 3/4 two whole times, and there is a remainder of 1/12. 1/12 is actually 1/4 of a serving, so our answer is that you ate 2 and 1/4 servings. Instead, we could write this division problem as a fraction of fractions, and then simplify it.  With fractions, we can multiply the numerator and denominator by the same thing without changing the value of the overall fraction.  In this case, we multiply the top and bottom by 3: $\dfrac{\quad\frac{3}{4}\quad}{\frac{1}{3}} = \dfrac{3 \cdot \frac{3}{4}}{3 \cdot \frac{1}{3}} = \dfrac{\frac{9}{4}}{1} = \dfrac{9}{4}$ Thus, our answer is that you ate  $\boxed{\dfrac{9}{4}}$ servings. Notice that 9/4 is equal to 2 and 1/4.   Click here to show solution 2. If 1/3 of a lasagna is equal to 1 serving, then a whole lasagna must be equal to 3 servings. You ate 3/4 of a whole lasagna, so you ate 3/4 of 3 servings. Often in math, the word “of” means multiplication. Thus, we can calculate that you ate: $\dfrac{3}{4} \cdot 3 \text{ servings}$ Writing the 3 as a fraction, we can multiply across to get: $\dfrac{3}{4} \cdot \dfrac{3}{1} \text{ servings} \quad \longrightarrow \quad \boxed{\dfrac{9}{4} \text{ servings}}.$ Recall that the fraction 9/4 can be thought of as 4/4 + 4/4 + 1/4 which means 2 and 1/4. As a decimal, we could write this as:  2.25 You ate more than 2 servings.  You probably ate more than you should’ve. Need Help ? MathPlanet – Rates and Ratios ARBs – Dividing One Fraction By Another

Problem 10:  Explain in words why the following properties of exponents make sense.

 Recall the meaning of an exponent:  ax  means  x  copies of  a  multiplied together. Use this to explain in words why the following properties of exponents make sense. Need Help ? MathPlanet – Exponent Rules MathPlanet – Properties of Exponents Part a. $a^{x} \cdot a^{y}=a^{x+y}$ Click here to show solution. If you are multiplying x copies of a and y copies of a all together, then in total you will be multiplying x+y copies of a. $a^{x} \cdot a^{y} = \underbrace{a \cdot a \cdot ... \cdot a}_\text{x copies of a} \cdot \underbrace{a \cdot a \cdot ... \cdot a}_\text{y copies of a}$ Part b. $\dfrac{a^{x}}{a^{y}}=a^{x-y}$ Click here to show solution. Division is the inverse of multiplication, so if you are multiplying x copies of a, and then dividing by y copies of a, then some of those copies will cancel out, and you will have x-y copies left over. $\dfrac{a^{x}}{a^{y}} = \dfrac{\overbrace{a \cdot a \cdot ... \cdot a}^\text{x copies of a} }{ \underbrace{a \cdot a \cdot ... \cdot a}_\text{y copies of a} }$ Part c. $\left(a^{x}\right)^{y}=a^{xy}$ Click here to show solution. If you are multiplying x copies of a, and then you take the result and multiply y copies of that, then you are really multiplying xy copies of a. $\left(a^{x}\right)^{y} = \underbrace{ ( \overbrace{a \cdot a \cdot ... \cdot a}^\text{x copies of a} ) \cdot ( \overbrace{a \cdot a \cdot ... \cdot a}^\text{x copies of a} ) \cdot ... \cdot ( \overbrace{a \cdot a \cdot ... \cdot a}^\text{x copies of a} )}_\text{y sets of parentheses}$

Problem 11:  Verify that each statement is false and then correct it.

 A student has simplified the following expressions incorrectly. First, substitute numbers in for the variables and verify that the expressions are not equivalent. Then, identify the student’s error and correct it by properly simplifying the original expression. Need Help ? Monterey – Distributive Property BrownMath – More Distributive Properties Part a. Click here to show solution. Let’s try a = 2 and b = 1: $(2a-3b)^{2} = (2(2)-3(1))^{2} = (4-3)^{2} = (1)^{2} = 2$ $4a^{2}-9b^{2} = 4(2)^{2} - 9(1)^{2} = 4(4) - 9(1) = 16 - 9 = 7$ So clearly there is some mistake, because they were supposed to be equal. The mistake was that the student distributed an exponent to each number and variable inside the parentheses.  We know that you can distribute an exponent over multiplication, but not subtraction, so we need to simplify this another way. Since squaring a quantity means multiplying it by itself, let’s try: $(2a-3b)^{2} = (2a-3b)(2a-3b)$ Now we can use FOIL, and expand: $(2a-3b)(2a-3b) = 4a^{2} -6ab -6ab +9b^{2}$ Notice that the last term is positive, because we multiplied two negative quantities together.  We can combine the middle terms, because they are like terms, to get: $4a^{2} -6ab -6ab +9b^{2} = 4a^{2} -12ab +9b^{2}$ Now we can see that the student was missing the middle term, -12ab, and also that the last term should be positive. Part b. Click here to show solution. Let’s try setting x = 2: $3x(x - 2) - 4(5 - x) = 3(2)(2 - 2) - 4(5 - 2) = -12$ $3x^{2} - 7x - 20 = 3(2)^{2} - 7(2) - 20 = 12 - 14 - 20 = -22$ Since these did not come out equal, there must be some error. Order of operations tells us that to simplify the original expression, we need to distribute to get rid of each set of parentheses. We can do this to get: $3x(x - 2) - 4(5 - x) = 3x \cdot x - 3x \cdot 2 - [ 4 \cdot 5 - 4 \cdot x ]$ Notice that we added brackets to make sure that the minus sign in front of the four is also distributed. $3x \cdot x - 3x \cdot 2 - [ 4 \cdot 5 - 4 \cdot x ] = 3x^{2} - 6x - [ 20 - 4x ]$ Now we can distribute the minus sign to get: $3x^{2} - 6x - [ 20 - 4x ] = 3x^{2} - 6x - 20 + 4x$ Notice that the 4x came out positive because the two minus signs cancelled.  Compare this to the student’s work, and see that they incorrectly found this term to be -x, and it looks like they simply forgot to distribute the 4 all the way. Combining like terms, we get:  $3x^{2} - 2x - 20$. Part c. Click here to show solution. Let’s try x = 3: $3(2x - 5)^{2} = 3(2(3) - 5)^{2} = 3(6 - 5)^{2} = 3(1)^{2} = 3(1) = 3$ $(6x - 15)^{2} = (6(3) - 15)^{2} = (18 - 15)^{2} = (3)^{2} = 9$ Before we even test the last step, we can see that the student has made an error, because the first expression did not come out equal to the second. Looking at the student’s work, we can see that they distributed the 3 into the parentheses, even though order of operations tells us that the exponent needs to be addressed first, and then we can distribute the 3. Following the proper order, we need to rewrite the squared part as: $3(2x - 5)^{2} = 3(2x - 5)(2x - 5)$ Now we need to use FOIL to combine the two sets of parentheses into one: $3(2x - 5)(2x - 5) = 3(4x^{2} - 10x - 10x + 25)$ Notice that we still have one big set of parentheses keeping the 3 separate from the rest.  This is very important, because the 3 must be distributed to every part: $3(4x^{2} - 10x - 10x + 25) = 12x^{2} - 30x - 30x + 75$ Finally, we can combine like terms to get:  $12x^{2} - 60x + 75$

Problem 12: Expand the following expressions. Remember to use FOIL.

 F.O.I.L. stands for First, Outer, Inner, Last. When two binomials are being multiplied together, we can use FOIL to remember all the products that appear in the expansion. For example:   $(a + b)(c + d) = ac + ad + bc + bd$ $\boxed{ac}$  is the product that results from multiplying the First terms together. $\boxed{ad}$  is the product of the Outer terms. $\boxed{bc}$  is the product of the Inner terms. $\boxed{bd}$  is the product of the Last terms. Need Help ? MathPlanet – Multiplication of Polynomials Part a. $(x+4)(x+6)$ Click here to show solution. Following FOIL will give us the following: $x \cdot x + x \cdot 6 + 4 \cdot x + 4 \cdot 6$ We can take each product and simplify it as: $x^{2} + 6x + 4x + 24$ Now we can combine 6x and 4x to make 10x: $\boxed{x^{2} + 10x + 24}.$ Part b. $(x+8)(x-3)$ Click here to show solution. Following FOIL will give us the following: $x \cdot x + x \cdot -3 + 8 \cdot x + 8 \cdot -3$ We can take each product and simplify it as: $x^{2} - 3x + 8x - 24$ Now we can combine -3x and 8x to make 5x: $\boxed{x^{2} + 5x - 24}.$ Part c. $(-x+2)(x-7)$ Click here to show solution. Following FOIL will give us the following: $-x \cdot x + -x \cdot -7 + 2 \cdot x + 2 \cdot -7$ We can take each product and simplify it as: $-x^{2} + 7x + 2x - 14$ Notice how -x times -7 gave us a positive result, because the negatives cancel each other. Now we can combine 7x and 2x to make 9x: $\boxed{-x^{2} + 9x - 14}.$ Part d. $(-x-1)(-x+5)$ Click here to show solution. Following FOIL will give us the following: $-x \cdot -x + -x \cdot 5 + -1 \cdot -x + -1 \cdot 5$ We can take each product and simplify it as: $x^{2} - 5x + 1x - 5$ Notice how the first term is positive, because it is the product of two negatives.  This is also true of the third term, 1x.  Also, instead of writing 1x, we usually just write x, and the coefficient of 1 is implied Now we can combine -5x and x to make -4x: $\boxed{x^{2} - 4x - 5}.$

Problem 13:  Rewrite each expression in factored form.

 Part a. $x^{2}+8x+12$ Click here to show solution. List the factors of the constant term, 12: $\begin{array}{c|c} & \\\hline 1 & 12 \\2 & 6 \\3 & 4 \\-1 & -12 \\-2 & -6 \\-3 & -4 \end{array}$ Notice that we include pairs where both numbers are negative. We want to choose the pair that adds up to the coefficient of x, which is 8, so we pick 2 and 6. $x^{2}+8x+12 = \boxed{(x+2)(x+6)}.$ Need Help ? MathPlanet – Factoring Quadratics Part b. $x^{2}+9x+8$ Click here to show solution. List the factors of the constant term, 8: $\begin{array}{c|c} & \\\hline 1 & 8 \\2 & 4 \\-1 & -8 \\-2 & -4 \end{array}$ Notice that we include pairs where both numbers are negative. We want to choose the pair that adds up to the coefficient of x, which is 9, so we pick 1 and 8. $x^{2}+9x+8 = \boxed{(x+1)(x+8)}.$ Part c. $x^{2}+2x-15$ Click here to show solution. List the factors of the constant term, -15: $\begin{array}{c|c} & \\\hline 1 & -15 \\3 & -5 \\-1 & 15 \\-3 & 5 \end{array}$ Notice that we include each pair twice, once with the smaller number negative and once with the larger number negative. We want to choose the pair that adds up to the coefficient of x, which is 2, so we pick -3 and 5, because adding them together gives us -3 + 5 = 2. $x^{2}+2x-15 = \boxed{(x-3)(x+5)}.$ Part d. $x^{2}-x-42$ Click here to show solution. List the factors of the constant term, -42: $\begin{array}{c|c} & \\\hline 1 & -42 \\2 & -21 \\3 & -14 \\6 & -7 \\-1 & 42 \\-2 & 21 \\-3 & 14 \\-6 & 7 \end{array}$ Notice that we include each pair twice, once with the smaller number negative and once with the larger number negative. We want to choose the pair that adds up to the coefficient of x, which is -1, so we pick 6 and -7, because when we add them together we get 6 + -7 = -1. $x^{2}-x-42 = \boxed{(x+6)(x-7)}.$ Part e. $x^{2}-7x+12$ Click here to show solution. List the factors of the constant term, 12: $\begin{array}{c|c} & \\\hline 1 & 12 \\2 & 6 \\3 & 4 \\-1 & -12 \\-2 & -6 \\-3 & -4 \end{array}$ Notice that we include pairs where both numbers are negative. We want to choose the pair that adds up to the coefficient of x, which is -7, so we need to pick two negative numbers.  We pick -3 and -4 because adding them gives us -3 + -4 = -7 $x^{2}-7x+12 = \boxed{(x-3)(x-4)}.$ Part f. $x^{2}+3x-40$ Click here to show solution. List the factors of the constant term, -40: $\begin{array}{c|c} & \\\hline 1 & -40 \\2 & -20 \\4 & -10 \\5 & -8 \\-1 & 40 \\-2 & 20 \\-4 & 10 \\-5 & 8 \end{array}$ Notice that we include each pair twice, once with the smaller number negative and once with the larger number negative. We want to choose the pair that adds up to the coefficient of x, which is 3, so we pick -5 and 8, because when we add them together we get -5 + 8 = 3. $x^{2}+3x-40 = \boxed{(x-5)(x+8)}.$

Problem 14:  Use the area diagram to answer the questions.

 Part a. Using the diagram below, explain why (a + b)2 ≠ a2 + b2. What is (a + b)2 actually equal to ? Click here to show solution. First, notice that the area of the big square, with side length (a+b), must be equal to the combined area of all four smaller rectangles. The area of any square is equal to the side length squared, so the area of the big square must be (a + b)2. Two of the four smaller rectangles are actually squares, one has sides of length a and the other has sides of length b.  This means one square has area a2 and the other has area b2. The two remaining rectangles have one side of length a and one side of length b, which gives each an area of a times b, or ab. The combined area of all four smaller rectangles is given by adding up the areas, which means the total area in the figure is:  a2 + ab + ab + b2.  Combining the middle two terms gives us:  a2 + 2ab + b2. This total area must be equal to the area of the big square, which is (a + b)2, so setting these expressions equal gives us: (a + b)2 = a2 + 2ab + b2. Clearly it would be incorrect to say that (a + b)2 = a2 + b2  because this ignores the contributions of the two rectangles with area ab. Need Help ? KhanAcademy – Area Model for Multiplying Binomials

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