## Finding a Second Degree Equation from Its Solutions

This post about finding a second degree equation from its solutions is part of a series of posts to help you prepare for the Advanced Algebra and Functions part of the Accuplacer test.

## Question

x = 4 is the only solution to one of the equations below.

Which one?

A.  $\fn_phv {\color{DarkRed} \left ( 4x \right )^2=0}$

B.  $\fn_phv {\color{DarkRed} \left ( x-4 \right )^2=0}$

C.  $\fn_phv {\color{DarkRed} \left ( x+4 \right )^2=0}$

D.  $\fn_phv {\color{DarkRed} \left ( x-4 \right )\left (x+4 \right )=0}$

## Solution

### Note

For solving equations, the terms “root,” “solution,” “zero,” and “x-intercept” all mean pretty much the same thing, an answer to the problem. “Zero” and “x-intercept” come from the way equations are usually graphed for a problem: The equation is set equal to zero and then solved for the x-value that makes it all equal zero. On a graph, that x-value is an x-intercept.

### Back to the Question

The factor theorem says that if $\inline \fn_jvn x=c$  is a solution to $\inline \fn_jvn f(x)=0$, then $\inline \fn_jvn x-c$ must be a factor of $\inline \fn_jvn f(x)$. For example, if $\inline \fn_jvn x=4$ is a solution, $\inline \fn_jvn x-4$ must be a factor of $\inline \fn_jvn f(x)$.

For which of the answer choices is $\inline \fn_jvn x-4$ a factor? Choices B and D. What’s the right answer?

The other clue in the question is that $\inline \fn_jvn x=4$ is the only solution. The factor theorem says also that if $\inline \fn_jvn x-c$ is a factor of $\inline \fn_jvn f(x)$, then c must be a solution to $\inline \fn_jvn f(x)$. Note that the equation in answer choice D also contains another factor: $\inline \fn_jvn x+4$. If $\inline \fn_jvn x+4$  is a factor, then -4 is a solution. That would make $\inline \fn_jvn x=-4$ a second solution — but the question says there can be only one, so D cannot be the correct answer.

That means the correct answer is choice B.

The Accuplacer sample problem on which this one is based is #12 in the sample questions for the Accuplacer Advanced Algebra and Functions test.

## Question

This post about how to find a quadratic function’s equation from its graph is part of a series of posts to help you prepare for the Advanced Algebra and Functions part of the Accuplacer test.

This graph represents a polynomial function $\inline \fn_jvn y=g(x)$. Which of the following could be g‘s equation?

1. $\inline \fn_jvn g(x)=x^2-2x-3$
2. $\inline \fn_jvn g(x)=-x^2+2x+3$
3. $\inline \fn_jvn g(x)=(x-1)(x+3)$
4. $\inline \fn_jvn g(x)=-(x-1)^2-4$

## Solution

What can you tell from the graph?

You can read the coordinates of the x-intercepts, the y-intercept, and the turning point.

 x-intercepts (-1, 0), (3, 0) y-intercept (0, -3) turning point (1, -4)

The x-intercepts are clues to g‘s factors: The x-intercept (-1, 0) tells you that $\fn_phv x+1$ is a zero, and the x-intercept (3, 0) tells you that $\fn_phv x-3$ is a factor.

Sidebar: The factor theorem says that a polynomial that has a polynomial has a factor $\fn_phv x-c$ has a root at c, and a polynomial that has a root at c has a factor $\fn_phv x-c$. A root is the x-value at the x-intercept: for this example, -1 and 3. So the factors are $\fn_phv x+1$ and $\fn_phv x-3$.

Multiply those factors together and see what you get:

$\fn_jvn {\color{DarkBlue} (x+1)(x-3)=x^2-2x-3}$

The graph looks like a parabola, a second-degree curve. And $\fn_phv x^2-2x-3$ is second degree. So you may be done. How can you check? Use one of the two other points you can read from the graph. The y-intercept is (0, -3). Plug in 0 for x and see if the equation gives you -3, the y-intercept.

If

$\fn_jvn {\color{DarkBlue} g(x)=x^2-2x-3}$

then

$\fn_jvn {\color{DarkBlue} g(0)=0^2-2 \cdot 0-3}$

And as we saw from the graph, the y-intercept is (0, -3). Check. So answer choice #1 is the correct one.

That is one way to find a quadratic function’s equation from its graph.

Alternatively, since this question is multiple choice, you could try each answer choice. The easiest way to do that, I think, is to substitute 0 for and see if you get that $\inline&space;\fn_phv&space;g\left&space;(&space;0&space;\right&space;)=-3$, the y-intercept:

1. $\inline \fn_jvn g(x)=x^2-2x-3$.  Then $\inline \fn_jvn g(0)=-3$. This is the answer we found. It works.
2. $\inline \fn_jvn g(x)=-x^2+2x+3$$\inline \fn_jvn g(0)=3\neq -3$. Not this one.
3. $\inline \fn_jvn g(x)=(x-1)(x+3)$$\inline \fn_jvn g(0)=-3$.  Looks like this works.
4. $\inline \fn_jvn g(x)=-(x-1)^2-4$$\inline \fn_jvn g(0)=-5$. Does not work.

So we have two contenders, answer choice #1 and answer choice #3. That means we have to try at least one more point besides the y-intercept. We have already done that for answer choice #1; we know the x-intercepts work for it. Let’s see if $\inline \fn_jvn y=0$ at $\inline \fn_jvn x=-1$ for answer choice #3, $\inline \fn_jvn g(x)=(x-1)(x+3)$.

$\fn_jvn {\color{DarkBlue} g(-1)=(-1-1)(-1+3)=-2 \cdot 2 = -4 \neq 0}$

So answer choice #3 does not work after all, and answer choice #1 is the correct one.

For more about the end behavior of polynomial functions, see “End Behavior, Degree, and Leading Coefficient.”

The Accuplacer sample problem on which this one is based is #10 in the sample questions for the Accuplacer Advanced Algebra and Functions test.

## How to Factor a Quadratic

This post about how to factor a quadratic is part of a series of posts to help you prepare for the Advanced Algebra and Functions part of the Accuplacer test.

## Question

Factor completely:

$\inline&space;\fn_jvn&space;{\color{DarkBlue}&space;2x^2-4x-6}$

## Solution

The first step in factoring a polynomial (that’s an expression, like this one, that has multiples of powers of x added together) is to look for a common factor. Notice that every one of the terms (the pieces that get added together) is even. That means each term has a factor of 2. Factor that out:

$\fn_jvn {\color{DarkBlue} 2x^2-4x-6=2(x^2-2x-3)}$

Are we done? That depends on whether $\inline \fn_jvn x^2-2x-3$  can be factored.

Can it?

### Sidebar: Finding the Magic Numbers

How do you factor a quadratic trinomial whose leading coefficient is 1 — that is, whose $\inline \fn_jvn x^2$ term is not multiplied by a number (just 1, implicitly)?

Let’s reverse the process to see what’s going on. Think about how two first-degree binomials — that is, terms that look like $\fn_phv x+$ a number — multiply into a trinomial. To do that, consider two binomials, $\inline \fn_jvn x+m$ and $\inline \fn_jvn x+n$. Multiply those together to get

$\fn_jvn {\color{DarkBlue} (x+m)(x+n)=x^2+(m+n)x+mn}$

So how can you go in the other direction? That is, how can you get something like  $\fn_cm \left ( x+m \right )\left ( x+n \right )$ from something like $\fn_cm x^2+bx+c$?

Your clues to m and n are that when you multiply them together you get the constant term, the last term, of the expression getting factored, and that when you add them together you get the multiplier of $\fn_phv x$ in the middle term.

Now let’s take an example. Factor $\inline \fn_jvn x^2+2x-8$.

$\fn_jvn {\color{DarkBlue}x^2+(m+n)x+mn}$

$\fn_jvn {\color{DarkBlue}x^2+2x-8}$

So $\inline \fn_jvn m+n=2$ and $\inline \fn_jvn mn=-8$. Can you find two numbers like that? Start with the multiplication. Integer factor pairs of -8, and their sums, are:

 m n m + n 1 -8 -7 –1 8 7 2 -4 -2 –2 4 2

It looks like -2, 4 is the ticket, because those two numbers multiply to -8 and add to 2. Then the factoring is:

$\fn_jvn {\color{DarkBlue}x^2+2x-8=(x-2)(x+4)}$

### Back to the Problem

Let’s try that out on $\inline \fn_jvn x^2-2x-3$. See if you can factor the last term, -3, into two numbers that add up to the multiplier of x in the middle term, -2.

$\fn_jvn {\color{DarkBlue}x^2+(m+n)x+mn}$

$\fn_jvn {\color{DarkBlue}x^2-2x-3 }$

-3 has two pairs of integer factors: One is -1 and 3; the other is 1 and -3.

$\fn_jvn {\color{DarkBlue}-3=-3 \cdot 1}$

$\fn_jvn {\color{DarkBlue}-3+1=-2}$

So yes, there are two numbers that multiply to -3 and add to -2. Those numbers are -3 and 1. $\inline \fn_jvn x^2-2x-3$  factors into two binomials, of which one ends with -3 and one ends with 1:

$\fn_jvn {\color{DarkBlue}x^2-2x-3=(x-3)(x+1)}$

To check, you can multiply back.

Finally, don’t forget that 2 is a factor in the original expression. (I just started ignoring it, because it was not relevant to what we were doing at the moment.) So the final answer is

$\fn_jvn {\color{DarkBlue}2(x-3)(x+1)}$

The Accuplacer sample problem to which this is similar is multiple choice: You are given four possible sets of factors of a trinomial. If you don’t want to factor, you can multiply the factors in each answer choice until you find one that works.

I called this thing you factor a quadratic. That means it’s a polynomial whose highest-power term is second power. Many quadratics, like the ones considered above, have three terms; these may also be called trinomials, and you could call the problem above an example of how to factor trinomials.

Factoring is often the first step in solving a quadratic. For more about that, see “Solving a Quadratic.”

This question is similar to #7 in the sample questions for the Accuplacer Advanced Algebra and Functions test.

## Question

This post about solving quadratic equations is part of a series of posts to help you prepare for the Advanced Algebra and Functions part of the Accuplacer test.

Find the values of x that satisfy this equation:

$\fn_jvn {\color{Blue} x^2+2x-10=5}$

## Solution

It’s a quadratic. That means the first step to solving it is to get everything on the left-hand side, leaving 0 on the right. In this case, that means that you subtract 5 from each side:

$\fn_jvn {\color{Blue} x^2+2x-15=0}$

Now see if you can solve this quadratic by factoring: Look for two numbers that multiply to -15 and add up to 2. 5 and -3 fit the bill. So one factor is $\fn_cm {\color{Blue} x+5}$ and one is $\fn_cm {\color{Blue} x-3}$. Then

$\fn_jvn {\color{Blue} x^2+2x-15=\left ( x+5 \right )\left (x-3 \right )}$

And we said

$\fn_jvn {\color{Blue} x^2+2x-15=0}$

So

$\fn_jvn {\color{Blue} \left ( x+5 \right )\left (x-3 \right )=0}$

Two things multiplied together can equal 0 only when at least one of those things equals 0. So to find solutions to (+ 5)(– 3) = 0, set x + 5 equal to 0 and set x – 3 equal to 0. Then the factor theorem says that a polynomial that has a factor $\fn_phv x-c$ has a root at c, and a polynomial that has a root at c has a factor $\fn_phv x-c$:

 $\fn_jvn {\color{Blue} x+5=0}$$\fn_jvn {\color{Blue} x+5=0}$ $\fn_jvn {\color{Blue} x-3=0}$$\fn_jvn {\color{Blue} x-3=0}$ $\fn_jvn {\color{Blue} x=-5}$$\fn_jvn {\color{Blue} x=-5}$ $\fn_jvn {\color{Blue} x=3}$$\fn_jvn {\color{Blue} x=3}$

So two values of x satisfy the equation: x=-5, 3. To check this answer, substitute these values for x into the equation. If you get a true equation, your answer is correct.

The Accuplacer sample question to which this question is similar is multiple choice. If you are unable to solve by factoring, you can plug in answer choices until you find one that works.

Note: Not every quadratic can be solved by factoring. When you can’t factor, you can complete the square or use the quadratic formula. This post shows you how to complete the square. The steps from completing the square to solving the equation are left to you.

For more about solving a quadratic, see “How to Factor a Quadratic” and “How Can You Find a Quadratic’s Equation from Its Graph?”

The Accuplacer sample problem on which this one is based is #9 in the sample questions for the Accuplacer Advanced Algebra and Functions test.