<< Chapter < Page Chapter >> Page >

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

{ −1 }

Got questions? Get instant answers now!

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 0  (double solution), 4 , and −4.

Got questions? Get instant answers now!
Got questions? Get instant answers now!

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

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

Got questions? Get instant answers now!

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 3 , −3 , and −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).
Got questions? Get instant answers now!
Got questions? Get instant answers now!

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

The sequence is {1,-1,1-1.....} has
amit Reply
circular region of radious
Kainat Reply
how can we solve this problem
Joel Reply
Sin(A+B) = sinBcosA+cosBsinA
Eseka Reply
Prove it
Eseka
Please prove it
Eseka
hi
Joel
June needs 45 gallons of punch. 2 different coolers. Bigger cooler is 5 times as large as smaller cooler. How many gallons in each cooler?
Arleathia Reply
find the sum of 28th term of the AP 3+10+17+---------
Prince Reply
I think you should say "28 terms" instead of "28th term"
Vedant
if sequence sn is a such that sn>0 for all n and lim sn=0than prove that lim (s1 s2............ sn) ke hole power n =n
SANDESH Reply
write down the polynomial function with root 1/3,2,-3 with solution
Gift Reply
if A and B are subspaces of V prove that (A+B)/B=A/(A-B)
Pream Reply
write down the value of each of the following in surd form a)cos(-65°) b)sin(-180°)c)tan(225°)d)tan(135°)
Oroke Reply
Prove that (sinA/1-cosA - 1-cosA/sinA) (cosA/1-sinA - 1-sinA/cosA) = 4
kiruba Reply
what is the answer to dividing negative index
Morosi Reply
In a triangle ABC prove that. (b+c)cosA+(c+a)cosB+(a+b)cisC=a+b+c.
Shivam Reply
give me the waec 2019 questions
Aaron Reply
the polar co-ordinate of the point (-1, -1)
Sumit Reply
Practice Key Terms 5

Get the best Algebra and trigonometry course in your pocket!





Source:  OpenStax, Algebra and trigonometry. OpenStax CNX. Nov 14, 2016 Download for free at https://legacy.cnx.org/content/col11758/1.6
Google Play and the Google Play logo are trademarks of Google Inc.

Notification Switch

Would you like to follow the 'Algebra and trigonometry' conversation and receive update notifications?

Ask