TOP 10 MUST KNOW BARUCH CSTM 0120 FINAL PROBLEMS

1 Step 1

TABLE OF CONTENTS

Baruch College CSTM 0120 Final Overview

CSTM 0120 is an prealgebra course in Baruch College. This is a non-credit course containing those topics from intermediate algebra that are needed prior to taking MTH 1030, College Algebra.

To view the syllabus for Baruch College CSTM 0120 CLICK HERE

To get your hands on Baruch College CSTM 0120 Last Minute Cram Course, CLICK HERE

To view sample final exam questions for Baruch College CSTM 0120 CLICK HERE.

Baruch CSTM 0120 Final Topic 1. Factoring Binomials.

Many binomial problems from the CSTM 0120 final can be solved with the application of the difference of squares formula.

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

CSTM 0120, Sample Final F019A, Problem 1

TOP 10 MUST KNOW BARUCH CSTM 0120 FINAL PROBLEMS 1

Step 1.

Let’s first factor $x^2-9$ by using the formula above:

$x^2-9=(x-3)(x+3)$

Step 2.

$x^2-64x$ by taking the greatest common factor:

$x^2-64x=x\left(x-64\right)$

We have $4$ different factors including:

$x-3$
$x+3$
$x$
$x-64$

The answer is $D$ , since $x-8$ is not included the list of the factors above.

Baruch CSTM 0120 Final Topic 2. Factoring $\mathbit{a}\mathbit{x}^\mathbf{2}+\mathbit{bx}+\mathbit{c}$ .

If you know quadratic equation by heart, you should be able to get a large part of CSTM 0120 final test problems correct.

Let’s refresh our memory:

$ax^2+bx+c=a(x-x_1)(x-x_2)$ , where:

$x_{1,2}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

CSTM 0120, Sample Final F019A, Problem 2

TOP 10 MUST KNOW BARUCH CSTM 0120 FINAL PROBLEMS 2
We need to completely factor $315x^6+18x^5-9x^4$

Step 1.

Quickly realize that it is a trinomial and chances are we need to turn that into a quadratic equation:

$315x^6+18x^5-9x^4=x^4(315x^2+18x-9)$

We can also see that every term in the trinomial is a factor of $9$ .

$9x^4(35x^2+2x-1)$

Step 2.

Only pay attention to the trinomial now and let’s factor it using the quadratic formula:

$35x^2+2x-1$

$a=35$

$b=2$

$c=-1$

$x_{1,2}=\frac{-2\pm\sqrt{2^2-4\ast35\ast(-1)}}{2\ast35}$

$x_{1,2}=\frac{-2\pm\sqrt{4+140}}{2\ast35}$

$x_{1,2}=\frac{-2\pm\sqrt{144}}{70}$

$x_{1,2}=\frac{-2\pm12}{70}$

$x_1=\frac{-2+12}{70}=\frac{10}{70}=\frac{1}{7}$

$x_2=\frac{-2-12}{70}=\frac{-14}{70}=\frac{-1}{5}$

$35x^2+2x-1=35\ast\left(x-\frac{1}{7}\right)\left(x+\frac{1}{5}\right)=(7x-1)(5x+1)$

Step 3.

Let’s combine the first two steps together:

$315x^6+18x^5-9x^4=9x^4\left(7x-1\right)(5x+1)$

From the multiple choices presented, $\left(7x-1\right)$ is one of the factors.

The answer is $C$ .

Baruch CSTM 0120 Final Topic 3. Factoring $\mathbit{x}^\mathbf{2}+\mathbit{bx}+\mathbit{c}$ .

Factoring $x^2+bx+c$ is similar to factoring $x^2+bx+c$ from the previous problem.

Let’s refresh our memory:

$x^2+bx+c=(x-x_1)(x-x_2)$ , where:

$x_{1,2}=\frac{-b\pm\sqrt{b^2-4c}}{2}$

CSTM 0120, Sample Final F022A, Problem 2

TOP 10 MUST KNOW BARUCH CSTM 0120 FINAL PROBLEMS 3
We need to completely factor $x^2-2x-3$ and $x^2+4x-5$

Step 1.

Let’s factor $x^2-2x-3$

$x_{1,2}=\frac{-b\pm\sqrt{b^2-4c}}{2}$

$x_{1,2}=\frac{-(-2)\pm\sqrt{{(-2)}^2-4\ast(-3)}}{2}$

$x_{1,2}=\frac{2\pm\sqrt{4+12}}{2}$

$x_{1,2}=\frac{2\pm\sqrt{16}}{2}$

$x_{1,2}=\frac{2\pm4}{2}$

$x_1=\frac{2+4}{2}=\frac{6}{2}=3$

$x_2=\frac{2-4}{2}=\frac{-2}{2}=-1$

$x^2-2x-3=(x-3)(x+1)$

Step 2.

Let’s factor $x^2+4x-5$

$x_{1,2}=\frac{-4\pm\sqrt{4^2-4\ast(-5)}}{2}$

$x_{1,2}=\frac{-4\pm\sqrt{16+20}}{2}$

$x_{1,2}=\frac{-4\pm\sqrt{36}}{2}$

$x_{1,2}=\frac{-4\pm6}{2}$

$x_1=\frac{-4+6}{2}=\frac{2}{2}=1$

$x_2=\frac{-4-6}{2}=\frac{-10}{2}=-5$

$x^2+4x-5=(x+5)(x-1)$

We have the following factors:

$x-3$
$x+1$
$x+5$
$x-1$

Thus, the answer is $D$ .

Baruch CSTM 0120 Final Topic 4. Greatest Common Factor.

The trick behind greatest common factor problems is to find the greatest common factor for every term individually.

Step 1.

Factor the constant

Step 2.

Find the greatest common factor of the constants

Step 3.

Find the greatest common factor for the first variable $x$

Step 4.

Find the greatest common factor for the second variable $y$

CSTM 0120, Sample Final F022A, Problem 3

TOP 10 MUST KNOW BARUCH CSTM 0120 FINAL PROBLEMS 4
In this problem we need to find the greatest common factor for the expression ${18x}^8y^2-48x^3y^5$

Step 1.

Let’s factor the two constants first:

$18=2\ast9=2\ast3^2$

$48=2\ast24=2\ast2\ast12=2\ast2\ast2\ast6=2\ast2\ast2\ast2\ast3=2^4\ast3$

Step 2.

Let’s identify the greatest common factor for the constants:

$2\ast3^2$ and $2^4\ast3$

$2\ast3=6$ would represent the greatest common factor of the constant.

Step 3.

Let’s now find the greatest common factor for the $x-term=x^3$

Step 4.

Finally, let’s find the greatest common factor for the $y-term=y^2$

Step 5.

Combining greatest common factors of the constant, $x$ -term and $y$ -term we get:

$6x^3y^2$

The answer is $D$ .

Baruch CSTM 0120 Final Topic 5. Line Equations.

Given a point $(x_1,y_1)$ and a slope $m$ the following formula can be used to find the line equation:

$y-y_1=m(x-x_1)$

CSTM 0120, Sample Final F019, Problem 19

TOP 10 MUST KNOW BARUCH CSTM 0120 FINAL PROBLEMS 5
Problem 19 of the Sample Final F019 is a great example of a linear equation problem. We are given a point $(5,7)$ and a slope $m=-4$ .

Step 1.

Plug in the values into the formula:

$y-7=-4(x-5)$

Step 2.

Multiply everything out and move x-term and y-term to one side and everything else to the other side:

$y-7=-4x+20$

$y+4x=7+20$

$4x+y=27$

The answer is $B$ .

1 Step 1

Baruch CSTM 0120 Final Topic 6. Slope.

A linear equation is defined as: $y=mx+b$ . In such equations, the slope is $m$ .

CSTM 0120, Sample Final F019, Problem 18

TOP 10 MUST KNOW BARUCH CSTM 0120 FINAL PROBLEMS 6
In problem 18 of the Sample Final F019, we need to find the slope of the function $y=-3+4x$ .

Let’s reorganize the equation $y=4x-3$ .

It is very clear and apparent that the slope of such function is $m=4$ .

The answer is $B$ .

Baruch CSTM 0120 Final Topic 7. Adding Rational Expressions.

Rational expressions problems typically include the following steps:

Step 1.

Identify lowest common denominator

Step 2.

Turn every fraction to the lowest common denominator

Step 3.

Combine both fractions

Step 4.

Simplify

CSTM 0120, Sample Final F019, Problem 14

TOP 10 MUST KNOW BARUCH CSTM 0120 FINAL PROBLEMS 7
We need to simplify $\frac{7}{20u^7}+\frac{8}{45u^4}$

Step 1.

Lowest common denominator for $20u^7 and 45u^4$ :

$20u^7=2\ast10\ast u^7=2\ast2\ast5\ast u^7=2^2\ast5\ast u^7$

$45u^4=3\ast15\ast u^4=3\ast3\ast5\ast u^4=3^2\ast5\ast u^4$

Thus, the lowest common denominator is:

$2^2\ast3^2\ast5\ast u^7=4\ast9\ast5\ast u^7=180u^7$

Step 2.

Let’s convert denominators of both terms of the expression to $180u^7$ :

$\frac{7}{20u^7}=\frac{7}{20u^7}\ast\frac{9}{9}=\frac{63}{180u^7}$

$\frac{8}{45u^4}=\frac{8}{45u^4}\ast\frac{4u^3}{4u^3}=\frac{32u^3}{180u^7}$

Step 3.

Combine both fractions:

$\frac{63}{180u^7}+\frac{32u^3}{180u^7}=\frac{63+32u^3}{180u^7}$

Step 4.

The fraction cannot be simplified further.

The answer is $E$ .

Baruch CSTM 0120 Final Topic 8. Systems of Linear Equations: Solving by Substitution.

Step 1.

Solve for $x$ in terms of $y$ or solve for $y$ in terms of $x$

Step 2.

Plug the found expression into the second equation and solve for the other term.

Sounds pretty confusing, but let’s see how we can apply such method on the existing system of linear equations.

CSTM 0120, Sample Final F039, Problem 26

TOP 10 MUST KNOW BARUCH CSTM 0120 FINAL PROBLEMS 8
We need to solve the following system of equations by substitution

$\begin{matrix}9x-4y=-70\\6x+y=-32\\\end{matrix}$

Step 1.

Second equation is much better for substitution purposes. Let’s solve for y from the second equation:

$6x+y=-32\rightarrow y=-32-6x$

Step 2.

Let’s plug the y-term into the first equation:

$9x-4\ast(-32-6x)=-70$

$9x+128+24x=-70$

$33x+128=-70$

$33x=-70-128$

$33x=-198\rightarrow x=-6$

Step 3.

Let’s now plug $x=-6$ into $y=-32-6x$

$y=-32-6\ast(-6)$

$y=-32+36\rightarrow y=4$

The answer is $(-6,4)$ .

Baruch CSTM 0120 Final Topic 9. Systems of Linear Equations: Solving by Addition.

Step 1.

Multiply both equations by respective multiplier to be able to cancel out $x$ -term or $y$ -term

Step 2.

Solve for the remaining term

Step 3.

Plug back the remaining term to solve for the eliminated term

Again, sounds pretty confusing, but let’s see how we can apply such method on the existing system of linear equations.

CSTM 0120, Sample Final F039, Problem 25

TOP 10 MUST KNOW BARUCH CSTM 0120 FINAL PROBLEMS 9
We need to solve the following system of equations by addition:

$\begin{matrix}2x+9y=-37\\5x-7y=55\\\end{matrix}$

Step 1.

Let’s cancel the y-term. We can do so by multiplying the top equation by $7$ and the bottom equation by $9$ .

$\begin{matrix}2x+9y=-37&|\times7\\5x-7y=55&|\times9\\\end{matrix}$

$\begin{matrix}14x+63y=-259\\45x-63y=495\\\end{matrix}$

Step 2.

After adding both equations we get:

$14x+63y+45x-63y=-259+495$

$59x=236\rightarrow x=4$

Step 3.

Plug in $x=4$ into the top equation to solve for $y$

$2x+9y=-37$

$2\ast4+9y=-37$

$8+9y=-37$

$9y=-45\rightarrow y=-5$

The answer is $(4,-5)$ .

Baruch CSTM 0120 Final Topic 10. Simplifying Complex Fractions.

Step 1.

Factor every fraction

Step 2.

Combine fractions together

Step 3.

Simplify

CSTM 0120, Sample Final F019, Problem 12

TOP 10 MUST KNOW BARUCH CSTM 0120 FINAL PROBLEMS 10
Let’s simplify the following expression:

$\frac{5t+15}{3t+15}\times\frac{15t+75}{t+75}$

Step 1.

Let’s factor every single term of the expression:

$\frac{5(t+3)}{3(t+5)}\times\frac{15(t+5)}{t+75}$

Step 2 & Step 3.

Combining and simplifying the fraction can be done in one action here:

$\frac{5(t+3)}{3(t+5)}\times\frac{15(t+5)}{t+75}=$

$\frac{75(t+3)(t+5)}{3(t+5)(t+75)}=$

$\frac{75(t+3)}{3(t+75)}=$

$\frac{25(t+3)}{(t+75)}$

The answer is $B$ .

1 Step 1

Share this content

Baruch College CSTM 0120 Final Overview

Baruch CSTM 0120 Final Topic 1. Factoring Binomials.

CSTM 0120, Sample Final F019A, Problem 1

Step 1.

Step 2.

Baruch CSTM 0120 Final Topic 2. Factoring .

CSTM 0120, Sample Final F019A, Problem 2

Step 1.

Step 2.

Step 3.

Baruch CSTM 0120 Final Topic 3. Factoring .

CSTM 0120, Sample Final F022A, Problem 2

Step 1.

Step 2.

Baruch CSTM 0120 Final Topic 4. Greatest Common Factor.

Step 1.

Step 2.

Step 3.

Step 4.

CSTM 0120, Sample Final F022A, Problem 3

Step 1.

Step 2.

Step 3.

Step 4.

Step 5.

Baruch CSTM 0120 Final Topic 5. Line Equations.

CSTM 0120, Sample Final F019, Problem 19

Step 1.

Step 2.

Baruch CSTM 0120 Final Topic 6. Slope.

CSTM 0120, Sample Final F019, Problem 18

Baruch CSTM 0120 Final Topic 7. Adding Rational Expressions.

Step 1.

Step 2.

Step 3.

Step 4.

CSTM 0120, Sample Final F019, Problem 14

Step 1.

Step 2.

Step 3.

Step 4.

Baruch CSTM 0120 Final Topic 8. Systems of Linear Equations: Solving by Substitution.

Step 1.

Step 2.

CSTM 0120, Sample Final F039, Problem 26

Step 1.

Step 2.

Step 3.

Baruch CSTM 0120 Final Topic 9. Systems of Linear Equations: Solving by Addition.

Step 1.

Step 2.

Step 3.

CSTM 0120, Sample Final F039, Problem 25

Step 1.

Step 2.

Step 3.

Baruch CSTM 0120 Final Topic 10. Simplifying Complex Fractions.

Step 1.

Step 2.

Step 3.

CSTM 0120, Sample Final F019, Problem 12

Step 1.

Step 2 & Step 3.

Share this content

You Might Also Like

Defeat Baruch College MTH 2207 – Final Test SP17 In 40 Minutes Or Less

Master Baruch College MTH 2003 – Final Test SP17 In 30 Minutes Or LESS

Destroy Baruch College MTH 1030 – Final Test FA18 In 30 Minutes Or LESS

GET ALL OF OUR

VIDEO SOLUTIONS

FOR JUST $7.99/MONTH!

Baruch CSTM 0120 Final Topic 2. Factoring $\mathbit{a}\mathbit{x}^\mathbf{2}+\mathbit{bx}+\mathbit{c}$ .

Baruch CSTM 0120 Final Topic 3. Factoring $\mathbit{x}^\mathbf{2}+\mathbit{bx}+\mathbit{c}$ .