REVIEW/TUTORIAL FOR THE COLLEGE ALGEBRA PLACEMENT TEST
1.	Evaluate – x2 – 3x + 9 for x = 1
2.	Multiply and simplify:
3.	Simplify and factor: 5x– 2(y – x) – x
4.	Simplify:
5.	Simplify and write as a single fraction:
6.	Simplify as much as possible:
7.	Simplify by rationalizing the denominator:
8.	Multiply: (8x3 y)(–3x2y7)
9.	Find the slope of the line: 2x – 3y + 10 = 0
10.	Write as a single fraction in terms of
11.	Solve for all values of x:
12.	Solve for all values of x: = 5
13.	Multiply and simplify:
14.	Find the solution set for x: 2x + 1 > 3x + 7
15.	Solve for x: 2x2 + 3x – 2 = 0
16.	Solve for x: 2x2 – 3x – 4 = 0
17.	Solve for x: 2x2 – 6x + 5 = 0
18.	Evaluate:
19.	Solve for x and y:
20.	Find the solution set for x:
21.	Write as a single fraction.
22.	Factor completely: x4 – 81.
23.	Add and simplify:
24.	The graphs of the system of equation consists of lines that
	a) intersect b) meet in one point c) are parallel
25.	Simplify:
26.	Find the solution set for x: x2 + 6 > 5x
27.	If f(x) = x3 + 1, find f(3)
28.	Graph: y = 1
29.	Graph: x = – 3
30.	Graph: 2x + 3y = 2

	ANSWERS FOR THE REVIEW/TUTORIAL
1.	Answer	5
		Evaluate – x2 – 3x + 9  for x = 1
		When substituting x = 1 in   – x2
		be sure to do the exponent before the multiplication by   – 1
		to get   – (1)2 = – 1.   So    – 1– 3 + 9 = 5
2.	Answer
		When multiplying so that
		becomes
3.	Answer	2(3x - y)
		Using the distribute law -2(y - x) = -2y + 2x we get:
		5x– 2(y – x) – x
		= 5x– 2y + 2x– x
		= 6x -2y
		= 2(3x - y)
4.	Answer
		To divide by 3, we invert and multiply
5.	Answer
		Write the numerator as a single fraction with denominator of 4,
		and write the denominator as a single fraction with denominator of 6,
		Now divide by
		Be sure to invert and multiply to get
		Here we divided the denominator 4 and the numerator 6 by 2.
6.	Answer
		,
		since x4 is positive. Similarly
7.	Answer
		Since we write
8.	Answer	–24x5y8
		When multiplying exponent expressions with the same base, keep the
		base and add the exponents thus, x3(x2) = x5 and y(y7) = y8
9.	Answer
		Write the equation of the line in the form y = mx + b or where m is the slope.
		3y = 2x – 10
10.	Answer
		Write and rationalize Now we are adding
11.	Answer
		Multiply both sides of the equation by the denominator x – 3
		Use the fact that
		1 – 3(x – 3) = x
		1 – 3x + 9 = x
		10 = 4x
12.	Answer	x = – 5,   x = 5
		There are two values of x whose absolute value is 5
		|-5| = 5   and   |5| = 5
13.	Answer
		We factor x2 – 3x = x(x– 3)   and   9 – x2 = (3 – x)(3 + x)
		Remember the difference of squares factorization a2 – b2 = (a – b)(a + b).
		Thus we get
		Use the fact that x – 3 = – (3 – x),   so that
		Putting this together we get
14.	Answer	x < – 6
		2x + 1 > 3x + 7
		Subtract 3x  from both sides – x + 1 > 7
		Subtract 1 from both sides – x > 6
		Multiply by – 1, we get x < – 6
		Remember multiplying an inequality by a minus changes the sense of the arrow.
15.	Answer	and x = – 2
		Consider the standard quadratic equation ax2 + bx + c = 0,
		whose solution is
		For the solution of 2x2 + 3x – 2 = 0, a = 2, b = 3, and c = – 2.
		Since the discriminant b2 – 4ac = 32 – 4(2)( –2) = 9 + 16 = 25 is a
		perfect square, we can factor directly: 2x2 + 3x – 2 = 0,
		(2x – 1)(x + 2) = 0
		2x – 1 = 0 x + 2 = 0
		x = – 2
16.	Answer
		For the solution of 2x2 – 3x – 4 = 0, we have a = 2, b = – 3, c = – 4.
		The discriminant b2 – 4ac = (– 3)2 – 4(2)( – 4) = 9 + 32 = 41 is not a
		perfect square so that we must use the quadratic formula and get
		two real roots: Thus
17.	Answer
		For the solution of 2x2 – 6x + 5 = 0, a = 2, b = – 6, c = 5
		The discriminant b2 – 4ac = (– 6)2 – 4(2)( 5) = 36 – 40 = – 4.
		Since the discriminant is negative, the two roots are imaginary.
		Thus, here we use
		We can reduce the answer by factoring 2 in the numerator
18.	Answer
		The negative exponent means we have to take the reciprocal of what is in
		the parentheses and then square.
		Remember when raising a fraction to a power, both the denominator and
		numerator are raised to that power
19.	Answer	x = 3, y = 2 Solve for x and y there are two methods that are often used.
		Addition method: Multiply the second equation by – 5
		5x – 3y = 21
		– 5x – 25y = 35
		Add the equations to get:
		– 28y = 56
		Substitute back into the second equation to get:
		x + 5(– 2) = – 7
		x – 10 = – 7
		x = –7 + 10 = 3
		Thus x = 3, y = 2.
		Substitution method: Solve for x in the second equation to get:
		x = –7 – 5y
		Substitute for x in the first equation to get:
		5(–7 – 5y) – 3y = 21   –35 – 25y – 3y = 21   –35 – 28y = 21 Add 35 – 28y = 21 + 35   – 28y = 56   Divide by – 28
		Substitute in equation two to get x. Substitute the value y = – 2
		in x = – 7– 5y to get x = –7– 5(–2) = – 7 + 10 = 3
		Thus x = 3, y = 2.
20.	Answer	Answer: 3 < x < 7
		We can do the problem algebraically or geometrically.
		Algebraically:
		|x - 2| < 5 means –5 < x– 2 < 5
		So by adding to 2 all three parts of the inequality we get
		–5 < x –2 < 5
		–5 + 2 < x –2 + 2 < 5 + 2
		–3 < x < 7
		Geometrically:
		|x - 2| < 5 means that the distance (in both directions) from x to 2 is less than 5.
		So if we move 5 units to the right (up the axis) from 2 we get 7 and 5 units
		to the left (down the axis) from 2 we get –3
21.	Answer
		The least common denominator is xy. Thus
		Since the denominators of the two fractions are now the same,
		we can add the numerators getting
22.	Answer	(x – 3)(x +3)(x2 + 9)
		Recall the factorization of the difference of squares a2 – b2 = (a – b)(a + b)
		x4 – 81 is a difference of squares namely (x2 )2 and 92.
		x4 – 81 = (x2 – 9)( x2 + 9). The first factor is again a difference of squares:
		(x2 – 9) = (x – 3)(x +3)
		x4 – 81 = (x2 – 9)( x2 + 9) = (x – 3)(x +3)(x2 + 9)
		x2 + 9, the sum of squares, does not factor
23.	Answer
		The least common denominator is 24y.
		Thus
		Here since the denominators are the same we add the numerator.
24.	Answer	The lines are parallel.
		If you solve for y in each equation to get the form y = mx + b,
		you can examine the slopes m and the y intercepts b.
		Equation 1 becomes: Equation 2 becomes:
		which reduces to
		Since the slopes are equal and the y intercepts are not, the lines are
		parallel. In the case that both the slopes and the y intercepts were equal,
		the lines would be the same.
		In the case that the slopes are unequal, the lines intersect in one point.
25.	Answer	2
		Think of as as
		is the cube root of 8 often written as
		So that
		the meaning of a negative exponent.
		is the square root of 4
		often written Thus
26.	Answer	x < 2  or  x > 3
		x2– 5x + 6 > 0
		To find the solution set for x set the inequality equal to zero and factor and solve for x:
		x2– 5x + 6 = 0
		(x – 3)(x – 2) = 0
		x = 3, x = 2
		Since the inequality is strictly greater than zero, neither of these are in the solution set.
		These two numbers divide the x axis into three sections:
		– ¥ < x < 2, 2 < x < 3, and 3 < x < ¥
		Pick any "test" number for each section and substitute into the inequality:
		For – ¥ < x < 2, say x = 0.
		Substituting x = 0 into x2– 5x + 6 gives us (0)2 – 5(0) + 6 = 6
		which is greater than zero.
		Thus – ¥ < x < 2 is part of the solution set.
		For 2 < x < 3, we can test with or
		Substituting into x2– 5x + 6  gives us
		which is less than zero.
		Thus 2 < x < 3, is NOT in our solution set.
		For 3 < x < ¥, we can substitute x = 4   to get
		42 – 5(4) +6 = 16 – 20 + 6 = 12 > 0.
		Thus 3 < x < ¥ is part of the solution set.
		Thus the complete solution set is {x| – ¥ < x < 2 or 3 < x < ¥}
		which can also be written in interval notation as (– ¥, 2) (3, ¥).
27.	Answer	28
		If f(x) = x3 + 1, f(3) = 33 + 1 = 28
28.	Answer	The horizontal line one unit above the x axis.
29.	Answer	The vertical line 3 units to the left of the y axis.
30.	Answer	The graph of  2x + 3y = 2   is
		To graph 2x + 3y = 2, we need to find any two points which lie on the line
		and connect them with the straight edge. There are several ways to do this.
		We outline two common methods.
		Intercept method:
		If we let x = 0, the y intercept is 3y = 2 or
		and if we let y =0 the x intercept is (1,0)
		Slope-Intercept method:
		Solving for y in the form y = mx + b we get:
		From this form we see that the y intercept is
		From this point we can use the slope, which is
		to find a second point by moving, 2 units to the left (0 – 2 = – 2)
		and 3 units up to get the second point
		Close