Welcome to our community

Be a part of something great, join today!

Algebra practice exam questions

  • Thread starter
  • Admin
  • #1


Staff member
Feb 24, 2012
A student I help has been given a set of questions to study for an upcoming exam, and has asked me to work the problems so that they can see the solutions worked out. I decided to post them here so that they may benefit others studying the same material.

Here are some formulas to which I will be referring:

(1) The point-slope formula

A line passing through the point $\left(x_1,y_1 \right)$ and having slope $m$ is given by:

\(\displaystyle y-y_1=m\left(x-x_1 \right)\)

(2) Difference of squares factorization

\(\displaystyle a^2-b^2=(a+b)(a-b)\)

(3) Discriminant of a quadratic

For the general quadratic $ax^2+bx+c$, the discriminant $\Delta$ is given by:


The nature of the roots of the quadratic depends in the discriminant. If it is positive , then there are two distinct real roots, if it is zero there is one real (repeated) root, and if it is negative then there are two distinct complex conjugate roots.

If the discriminat is a perfect square, then the roots are rational, otherwise they are irrational.

(4) The quadratic formula

The roots of the general quadratic given in (3) are given by:

\(\displaystyle x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\)

(5) A consequence of the remainder theorem

If $f(x)$ is divisible by $x-r$, then $f(r)=0$.

(6) The distance formula

The distance $d$ between two points $\left(x_1,y_1 \right),\,\left(x_2,y_2 \right)$ in the plane is given by:

\(\displaystyle d=\sqrt{\left(x_2-x_1 \right)^2+\left(y_2-y_1 \right)^2}\)

(7) The midpoint formula

Given the two points in (6), the midpoint is then:

\(\displaystyle \left(x_M,y_M \right)=\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2} \right)\)

Non-Calculator Practice Exam

Question 1

Find the equation of a line passing through the point $(2,6)$ which is parallel to the line with equation $y=2x+1$ expressing it in the form $ax+by+c=0$.

We are given the point $(2,6)$ and the slope $m=2$, since parallel lines have equal slopes. Using the point-slope formula (1), we find that the line may be written as:

\(\displaystyle y-6=2(x-2)\)

\(\displaystyle y-6=2x-4\)

\(\displaystyle y=2x+2\)

Arranging the equation in the desired form, we have:

\(\displaystyle 2x-y+2=0\)

Question 2

Fully factorize each of the following:

a. \(\displaystyle (x+1)^2-25\)

Recognizing that $25$ is the square of $5$, we then have the difference of squares:

\(\displaystyle (x+1)^2-5^2\)

Using the difference of squares formula (2), we then have:

\(\displaystyle ((x+1)+5)((x+1)-5)=(x+6)(x-4)\)

b. \(\displaystyle x^2-4x+1=0\)

Observing that the discriminant (3) $\Delta=(-4)^2-4(1)(1)=12$ is not a perfect square, we know then that the roots are not rational, hence application of the quadratic formula (4) yields:

\(\displaystyle x=\frac{-(-4)\pm\sqrt{12}}{2(1)}=2\pm\sqrt{3}\)

and so we may state:

\(\displaystyle x^2-4x+1=(x-(2+\sqrt{3}))(x-(2-\sqrt{3}))=(x-2-\sqrt{3})(x-2+\sqrt{3})=0\)

c. \(\displaystyle x^3+8x^2+17x+10\), given that \(\displaystyle x+2\) is a factor.

From the given information, we may state:

\(\displaystyle x^3+8x^2+17x+10=(x+2)(x^2+kx+5)=x^3+(k+2)x^2+(5+2k)x+10\)

Equating coefficients, we then find that \(\displaystyle k=6\), hence:

\(\displaystyle x^3+8x^2+17x+10=(x+2)(x^2+6x+5)\)

Now, we may further factor the quadratic factor (since the discriminant is $16$) to get:

\(\displaystyle x^3+8x^2+17x+10=(x+2)(x+1)(x+5)\)

Note, we could also have used the rational roots theorem or polynomial/synthetic division as well. For more information on factoring quadratics, see this topic:


Question 3

For what values of $k$ does \(\displaystyle x^2+kx+4=0\) have one solution?

Recall that a quadratic has a repeated root iff the discriminant (3) is zero. Thus, we require:

\(\displaystyle k^2-4(1)(4)=0\)

\(\displaystyle k^2=16\)

\(\displaystyle k=\pm4\)

Question 4

The cubic \(\displaystyle f(x)=3x^3-(2k+3)x^2+30x-k\) is divisible by \(\displaystyle x-2\). Find the value of $k$.

By the remainder theorem (5), we know:

\(\displaystyle f(2)=0\)


\(\displaystyle 3(2)^3-(2k+3)(2)^2+30(2)-k=0\)

\(\displaystyle 3(8)-(2k+3)(4)+30(2)-k=0\)

\(\displaystyle 24-8k-12+60-k=0\)

\(\displaystyle 72=9k\)

\(\displaystyle k=8\)

Question 5

The graph \(\displaystyle y=\frac{a}{x-5}-1\) has a $y$-intercept of 1.

a. SHOW that $a=-10$

We are told that \(\displaystyle y(0)=1\) and so:

\(\displaystyle 1=\frac{a}{0-5}-1\)

\(\displaystyle 2=-\frac{a}{5}\)

\(\displaystyle 10=-a\,\therefore\,a=-10\)

b. State the equations of any asymptotes

There must be a vertical asymptote where division by zero occurs, and so equating the denominator of the first term to zero, we find:

\(\displaystyle x-5=0\,\therefore\,x=5\)

For values of $x$ having unbounded magnitude, i.e., for \(\displaystyle x\to\pm\infty\) the term \(\displaystyle \frac{a}{x-5}\to0\) and to the horizontal asymptote is $y=-1$.

c. Find the exact value of any $x$-intercepts.

Using $a=-10$ and $y=0$, we find:

\(\displaystyle 0=\frac{-10}{x-5}-1\)

\(\displaystyle 1=\frac{10}{5-x}\)

\(\displaystyle 5-x=10\)

\(\displaystyle x=-5\)

Thus, the $y$-intercept is $(-5,0)$.

Question 6

For the triangle in the plane having vertices $A(-1,1),\,B(4,2),\,C(1,4)$

a. Show here that it is an isosceles triangle by proving that the length of $\overline{AC}=\overline{BC}$. Lengths are to be expressed in simplest exact form.

Using the distance formula (6), we find:

\(\displaystyle \overline{AC}=\sqrt{(1+1)^2+(4-1)^2}=\sqrt{4+9}=\sqrt{13}\)

\(\displaystyle \overline{BC}=\sqrt{(1-4)^2+(4-2)^2}=\sqrt{9+3}=\sqrt{13}\)

b. If $M$ is the midpoint of $\overline{AB}$, show that $\overline{CM}$ is perpendicular to $\overline{AB}$.

Using the midpoint formula (7), we find point $M$ is:

\(\displaystyle \left(x_M,y_M \right)=\left(\frac{-1+4}{2},\frac{1+2}{2} \right)=\left(\frac{3}{2},\frac{3}{2} \right)\)

The slope of $\overline{CM}$ is \(\displaystyle \frac{4-\frac{3}{2}}{1-\frac{3}{2}}=-5\)

The slope of $\overline{AB}$ is \(\displaystyle \frac{2-1}{4+1}=\frac{1}{5}\)

Since the produce of these slopes is $-1$, we may conclude they are perpendicular.

CAS Calculator Enabled Practice

Question 1

The midpoint of the line between $(-1,5)$ and $(a,b)$ is $(2,2)$. Find the coordinates of the point $(a,b)$.

Using the midpoint formula (7), we must have:

\(\displaystyle \frac{-1+a}{2}=2\)

\(\displaystyle a-1=4\)

\(\displaystyle a=5\)

\(\displaystyle \frac{5+b}{2}=2\)

\(\displaystyle b+5=4\)

\(\displaystyle b=-1\)

Thus, we have $(a,b)=(5,-1)$.

Question 2

The line \(\displaystyle 5x-y+16=0\) may be written as:

\(\displaystyle y-6=5(x+2)\)

Hence, we know it has gradient $5$ and passes through the point $(-2,6)$.

Question 3

If the two lines $5x-2y+1=0$ and $kx+y-12=0$ are p[erpendicular, then find $k$.

The slope of the first line is \(\displaystyle \frac{5}{2}\) and for the second line is $-k$. We require their product to be $-1$, thus:

\(\displaystyle \frac{5}{2}(-k)=-1\)

\(\displaystyle k=\frac{2}{5}\)

Question 4

An expression for the discriminant of \(\displaystyle mx^2-2mx-4\) is:

\(\displaystyle \Delta=(-2m)^2-4(m)(-4)=4m^2+16m=4m(m+4)\)

Question 5

Find a possible equation for the graph shown:


Using the roots, we may state:

\(\displaystyle f(x)=k(x+2)(x-1)(x-3)\) where $k<0$ since \(\displaystyle \lim_{x\to\pm\infty}=\mp\infty\). If we let $k=-1$, then we could write:

\(\displaystyle f(x)=(x+2)(x-1)(3-x)\)


Well-known member
Dec 25, 2012
Was this for a specific class, or for all of algebra in general?
  • Thread starter
  • Admin
  • #3


Staff member
Feb 24, 2012
Was this for a specific class, or for all of algebra in general?
It was a specific class, but I don't recall the title of the class. The student is in high school, and I know basic calculus was also being taught in this class as well.


Well-known member
Dec 25, 2012
Awesome! I was curious because I have been helping someone practice for the math section of the ACT :D