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Solve: ( x + 5 ) 3 2 = 8.

{ −1 }

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Solving equations using factoring

We have used factoring to solve quadratic equations, but it is a technique that we can use with many types of polynomial equations, which are equations that contain a string of terms including numerical coefficients and variables. When we are faced with an equation containing polynomials of degree higher than 2, we can often solve them by factoring.

Polynomial equations

A polynomial of degree n is an expression of the type

a n x n + a n 1 x n 1 + + a 2 x 2 + a 1 x + a 0

where n is a positive integer and a n , , a 0 are real numbers and a n 0.

Setting the polynomial equal to zero gives a polynomial equation    . The total number of solutions (real and complex) to a polynomial equation is equal to the highest exponent n .

Solving a polynomial by factoring

Solve the polynomial by factoring: 5 x 4 = 80 x 2 .

First, set the equation equal to zero. Then factor out what is common to both terms, the GCF.

5 x 4 80 x 2 = 0 5 x 2 ( x 2 16 ) = 0

Notice that we have the difference of squares in the factor x 2 16 , which we will continue to factor and obtain two solutions. The first term, 5 x 2 , generates, technically, two solutions as the exponent is 2, but they are the same solution.

5 x 2 = 0 x = 0 x 2 16 = 0 ( x 4 ) ( x + 4 ) = 0 x = 4 x = −4

The solutions are x = 0  (double solution), x = 4 , and x = −4.

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Solve by factoring: 12 x 4 = 3 x 2 .

x = 0 , x = 1 2 , x = 1 2

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Solve a polynomial by grouping

Solve a polynomial by grouping: x 3 + x 2 9 x 9 = 0.

This polynomial consists of 4 terms, which we can solve by grouping. Grouping procedures require factoring the first two terms and then factoring the last two terms. If the factors in the parentheses are identical, we can continue the process and solve, unless more factoring is suggested.

x 3 + x 2 9 x 9 = 0 x 2 ( x + 1 ) 9 ( x + 1 ) = 0 ( x 2 9 ) ( x + 1 ) = 0

The grouping process ends here, as we can factor x 2 9 using the difference of squares formula.

( x 2 9 ) ( x + 1 ) = 0 ( x 3 ) ( x + 3 ) ( x + 1 ) = 0 x = 3 x = −3 x = −1

The solutions are x = 3 , x = −3 , and x = −1. Note that the highest exponent is 3 and we obtained 3 solutions. We can see the solutions, the x- intercepts, on the graph in [link] .

Coordinate plane with the x-axis ranging from negative 5 to 5 and the y-axis ranging from negative 30 to 20 in intervals of 5. The function x cubed plus x squared minus nine times x minus nine equals zero is graphed along with the points (negative 3,0), (negative 1,0), and (3,0).
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Solving radical equations

Radical equations are equations that contain variables in the radicand    (the expression under a radical symbol), such as

3 x + 18 = x x + 3 = x 3 x + 5 x 3 = 2

Radical equations may have one or more radical terms, and are solved by eliminating each radical, one at a time. We have to be careful when solving radical equations, as it is not unusual to find extraneous solutions    , roots that are not, in fact, solutions to the equation. These solutions are not due to a mistake in the solving method, but result from the process of raising both sides of an equation to a power. However, checking each answer in the original equation will confirm the true solutions.

Radical equations

An equation containing terms with a variable in the radicand is called a radical equation    .

Given a radical equation, solve it.

  1. Isolate the radical expression on one side of the equal sign. Put all remaining terms on the other side.
  2. If the radical is a square root, then square both sides of the equation. If it is a cube root, then raise both sides of the equation to the third power. In other words, for an n th root radical, raise both sides to the n th power. Doing so eliminates the radical symbol.
  3. Solve the remaining equation.
  4. If a radical term still remains, repeat steps 1–2.
  5. Confirm solutions by substituting them into the original equation.

Questions & Answers

An investment account was opened with an initial deposit of $9,600 and earns 7.4% interest, compounded continuously. How much will the account be worth after 15 years?
Kala Reply
lim x to infinity e^1-e^-1/log(1+x)
given eccentricity and a point find the equiation
Moses Reply
12, 17, 22.... 25th term
Alexandra Reply
12, 17, 22.... 25th term
Akash
College algebra is really hard?
Shirleen Reply
Absolutely, for me. My problems with math started in First grade...involving a nun Sister Anastasia, bad vision, talking & getting expelled from Catholic school. When it comes to math I just can't focus and all I can hear is our family silverware banging and clanging on the pink Formica table.
Carole
find the 15th term of the geometric sequince whose first is 18 and last term of 387
Jerwin Reply
I know this work
salma
The given of f(x=x-2. then what is the value of this f(3) 5f(x+1)
virgelyn Reply
hmm well what is the answer
Abhi
how do they get the third part x = (32)5/4
kinnecy Reply
can someone help me with some logarithmic and exponential equations.
Jeffrey Reply
sure. what is your question?
ninjadapaul
20/(×-6^2)
Salomon
okay, so you have 6 raised to the power of 2. what is that part of your answer
ninjadapaul
I don't understand what the A with approx sign and the boxed x mean
ninjadapaul
it think it's written 20/(X-6)^2 so it's 20 divided by X-6 squared
Salomon
I'm not sure why it wrote it the other way
Salomon
I got X =-6
Salomon
ok. so take the square root of both sides, now you have plus or minus the square root of 20= x-6
ninjadapaul
oops. ignore that.
ninjadapaul
so you not have an equal sign anywhere in the original equation?
ninjadapaul
hmm
Abhi
is it a question of log
Abhi
🤔.
Abhi
I rally confuse this number And equations too I need exactly help
salma
But this is not salma it's Faiza live in lousvile Ky I garbage this so I am going collage with JCTC that the of the collage thank you my friends
salma
Commplementary angles
Idrissa Reply
hello
Sherica
im all ears I need to learn
Sherica
right! what he said ⤴⤴⤴
Tamia
hii
Uday
hi
salma
what is a good calculator for all algebra; would a Casio fx 260 work with all algebra equations? please name the cheapest, thanks.
Kevin Reply
a perfect square v²+2v+_
Dearan Reply
kkk nice
Abdirahman Reply
algebra 2 Inequalities:If equation 2 = 0 it is an open set?
Kim Reply
or infinite solutions?
Kim
The answer is neither. The function, 2 = 0 cannot exist. Hence, the function is undefined.
Al
y=10×
Embra Reply
Practice Key Terms 5

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Source:  OpenStax, College algebra. OpenStax CNX. Feb 06, 2015 Download for free at https://legacy.cnx.org/content/col11759/1.3
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