Easy Factorization Question

In summary, the given expression is factored by grouping the first two terms and the last two terms together, resulting in (x^2 - 4)(x - 2). This is because the first two terms have a common factor of (x - 2), and the last two terms have a common factor of (x - 2) as well. This can also be seen through the distributive property, where (x^2 - 4)(x - 2) is equivalent to x^2(x - 2) - 4(x - 2). The exponent symbol (^) indicates that the number or variable before it is raised to the power of the number or variable after it.
  • #1
77
0
Factorize:
x^3 − 2x^2 − 4x + 8 correct answer:

(x^3 − 2x^2) − (4x − 8)

x^2(x − 2) − 4(x − 2)

(x^2 − 4)(x − 2)

(x − 2)(x + 2)(x − 2)

(x − 2) 2(x + 2)

In the third line where the terms are grouped i don't understand why one of the (x - 2) is omitted? i.e shouldn't it be (x -2)^2 ?
 
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  • #2
Sorry just realized this should be under the "homework" section. Will post homework questions there in future (in the meantime could a mod please move it there)..
 
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  • #3
can anyone help please? Its too critical and basic to go any further in the course notes until i understand this
 
  • #4
... What is ^?
 
  • #5
alpha01 said:
can anyone help please? Its too critical and basic to go any further in the course notes until i understand this

(x^2 - 4) = (x - 2)(x + 2), so one of the (x - 2) factors was "hiding inside" the (x^2 - 4).
 
  • #6
Going from:x^2(x − 2) − 4(x − 2)
to:(x^2 − 4)(x − 2)
is your problem? Well consider this:
ab - cb = b(a - c)
Do you agree? If we multiply out on the right we get the left-hand side. Now in your example, a = x^2, b=(x-2), c=4.
Do you see how it works now?
 
  • #7
qspeechc said:
Going from:x^2(x − 2) − 4(x − 2)
to:(x^2 − 4)(x − 2)
is your problem? Well consider this:
ab - cb = b(a - c)
Do you agree? If we multiply out on the right we get the left-hand side. Now in your example, a = x^2, b=(x-2), c=4.
Do you see how it works now?



thank you
 

1. How do you factorize a polynomial equation?

To factorize a polynomial equation, you can use a variety of methods such as the grouping method, the difference of squares method, or the quadratic formula. It is important to first identify any common factors and then use the appropriate method to factorize the remaining terms.

2. Can all polynomial equations be factorized?

No, not all polynomial equations can be factorized. Some equations, such as prime polynomials, cannot be factored further. However, most polynomial equations can be factorized using various methods.

3. Why is factorization important in mathematics?

Factorization is important in mathematics because it helps us simplify complex expressions and solve equations. It also allows us to find the roots or solutions of an equation, which is useful in many real-life applications.

4. What is the difference between factorization and prime factorization?

Factorization is the process of breaking down a number or expression into smaller factors. Prime factorization, on the other hand, is the process of breaking down a number or expression into its prime factors. In other words, prime factorization is a more specific type of factorization.

5. How can factorization be used in real life?

Factorization has many real-life applications, such as in cryptography, where it is used to encrypt and decrypt messages. It is also used in economics to find the most efficient way to allocate resources. In engineering, factorization is used to simplify and solve complex equations. Additionally, it is used in statistics to find the prime factors of large numbers, which is useful in data analysis.

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