Some factorizing and other equations,

  • Thread starter Saturnfirefly
  • Start date
In summary, the two equations are not the same. One is -9x2 + 3x + 12, while the other is 3x2 - 9x + 12. The first equation can be factored out to many different integers, but the second equation cannot.
  • #1
Saturnfirefly
11
0

Homework Statement


1, 2. Factorize completely:
80t^5-5t

3X-9X^2+12

solve: 3. 1/2x = 1/6+1/3X

4. A right-angle triangle has the hypotenuse 16.0, one of the short sides is 3 cm longer than the other. Area and perimeter?

Homework Equations





The Attempt at a Solution


1. 5t(16t^4-1)
2. x(3-9x)+12 are they completely factorized though?
3. I cross multiplied until I got: 18X = 5X^2 + 12X... can't get any further
4. no success
 
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  • #2
Saturnfirefly said:

Homework Statement


1, 2. Factorize completely:
80t^5-5t

3X-9X^2+12

solve: 3. 1/2x = 1/6+1/3X

4. A right-angle triangle has the hypotenuse 16.0, one of the short sides is 3 cm longer than the other. Area and perimeter?

Homework Equations





The Attempt at a Solution


1. 5t(16t^4-1)
Notice that 16t2= (4t)2 and 1= 12. There is a special formula for a "difference of two squares".

2. x(3-9x)+12 are they completely factorized though?
That isn't "factored" at all! It is not ( )( ). You should be able to see immediately that you can factor out a 3: 3x2- 9x+ 12= 3(x2- 3x+ 4). Now can you factor 4 into two integers whose "sum" is -3?

3. I cross multiplied until I got: 18X = 5X^2 + 12X... can't get any further
You mean, I think, "solve 1/(2x)= 1/6+ 1/(3x). (What you wrote could be interpreted as (1/2)x= 1/6+ (1/3)x.) You didn't really need to multiply the left side by 18: 6 would have done it. 1/6+ 1/(3x)= 1/6+ 2/(6x)= (x+ 2)/6x Any way, "cross multiplying": multiply the left by 18x and the right by 2x gives 18x= 2x2+ 4x. I find it confusing to "cross multiply". I would prefer to say "multiply both sides by the least common denominator" which is 6x: 6x(1/(2x))= 6x(1/6)+ (6x)(1/3x) so 3= x+ 2 (not what I got above- I messed up the "cross multiply"!) . Can you solve 3= x+ 2?

4. no success
A right-angle triangle has the hypotenuse 16.0, one of the short sides is 3 cm longer than the other. Area and perimeter?
Well, a "relevant equation" would be the Pythagorean theorem. If we call one side "x" and the other side is "3 cm longer", what would the other side be? Now put those into the Pythagorean theorem to get the equation you need to solve. Once you know the lengths of the sides, it is easy to find the area and perimeter of the triangle.
 
  • #3
HallsofIvy said:
Notice that 16t2= (4t)2 and 1= 12. There is a special formula for a "difference of two squares".


That isn't "factored" at all! It is not ( )( ). You should be able to see immediately that you can factor out a 3: 3x2- 9x+ 12= 3(x2- 3x+ 4). Now can you factor 4 into two integers whose "sum" is -3?


You mean, I think, "solve 1/(2x)= 1/6+ 1/(3x). (What you wrote could be interpreted as (1/2)x= 1/6+ (1/3)x.) You didn't really need to multiply the left side by 18: 6 would have done it. 1/6+ 1/(3x)= 1/6+ 2/(6x)= (x+ 2)/6x Any way, "cross multiplying": multiply the left by 18x and the right by 2x gives 18x= 2x2+ 4x. I find it confusing to "cross multiply". I would prefer to say "multiply both sides by the least common denominator" which is 6x: 6x(1/(2x))= 6x(1/6)+ (6x)(1/3x) so 3= x+ 2 (not what I got above- I messed up the "cross multiply"!) . Can you solve 3= x+ 2?

I've solved the last one. But the two first I can't get at all :(...
 
  • #4
The first one: 5t(16t4-1) can be further factorized by a great amount. And remember:
(a - b)(a + b) = a2 - ab + ab - b2 = a2 - b2

For the next one: (x + a)(x +b) = x2 + bx + ax + ab = x2 + x(a + b) + ab

Try to compare your equations with these.
 
  • #5
What are the possible ways of factoring 4 into two integers? Do any of those add up to -3?
 
  • #6
For the second question, the original equation is -9x2 + 3x + 12, not 3x2 - 9x + 12, but HallsofIvy's suggestions for factoring still apply.
 

1. What is factorizing?

Factorizing is the process of finding the factors, or smaller numbers that can be multiplied together to get a larger number, of a given expression or equation. This is commonly used in algebra and can help simplify complicated equations.

2. How do you factorize an equation?

To factorize an equation, you need to identify the common factors between all the terms in the equation. You can then use the distributive property to rewrite the equation in a factored form. It may also be helpful to use techniques such as grouping or the difference of squares to factorize more complex equations.

3. What are some common methods for solving equations?

Some common methods for solving equations include factoring, substitution, elimination, and graphing. Each method may be more suitable for different types of equations, so it is important to understand the strengths and weaknesses of each method.

4. Can all equations be solved using factorizing?

No, not all equations can be solved using factorizing. Some equations may require more advanced methods such as quadratic formula or completing the square. It is important to understand when to use factorizing and when to use other methods for solving equations.

5. How can factorizing be used in real-life situations?

Factorizing can be used in various real-life situations, such as in finance to find the most efficient way to distribute money between different investments, or in biology to understand the genetic traits of offspring. It is a useful skill for problem-solving and critical thinking in many fields.

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