Some factorizing and other equations,

[Saturnfirefly]

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

 

[HallsofIvy]

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.

 

[Saturnfirefly]

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 :(...

 

[Jamil (2nd)]

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.

 

[HallsofIvy]

What are the possible ways of factoring 4 into two integers? Do any of those add up to -3?

 

[Tedjn]

For the second question, the original equation is -9x² + 3x + 12, not 3x² - 9x + 12, but HallsofIvy's suggestions for factoring still apply.