again, its (4x+3)^3
thanks
How do I factor this: (4x+3)^3 ?
2009-01-10 4:36 pm
回答 (10)
2009-01-10 4:45 pm
✔ 最佳答案
It is already factorized. You might want to do the opposite?(4x+3)^3 = (4x)^3 + 3(4x)^2 (3) + 3(4x)(3^2) + 3^3
Answer: 64x^3 + 144x^2 + 108x + 27
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Ideas: (a+b)^3 = a^3 + 3a^2 b + 3a b^2 + b^3
2009-01-11 12:39 am
Already factorized. Nothing to do.
2009-01-11 12:39 am
(4x+3)(4x+3)(4+3) FOIL the first two, and multiply that sum times the last one.
2009-01-11 12:57 am
(4x+3)^3
We know the formula (a+b)^3=a^3+b^3+3ab(a+b)
in the same way (4x+3)^3=64x^3+27+144x^2+108x (or)
64x^3+144x^2+108x+27
We know the formula (a+b)^3=a^3+b^3+3ab(a+b)
in the same way (4x+3)^3=64x^3+27+144x^2+108x (or)
64x^3+144x^2+108x+27
2009-01-11 12:55 am
You don't need to factor what is already factored. What you probably mean is FOIL. And the way to do that is in this manner.
(4x + 3)^3 expanded is:
(4x + 3)*(4x + 3)*(4x + 3)
(4x + 3)*(16x^2 + 12x + 12x + 9) = (4x + 3)*(16x^2 + 24x + 9)
64x^3 + 96x^2 + 36x + 48x^2 + 72x + 27
Now combine like terms:
64x^3 + 144x^2 + 108x + 27
There is your foiled product.
(4x + 3)^3 expanded is:
(4x + 3)*(4x + 3)*(4x + 3)
(4x + 3)*(16x^2 + 12x + 12x + 9) = (4x + 3)*(16x^2 + 24x + 9)
64x^3 + 96x^2 + 36x + 48x^2 + 72x + 27
Now combine like terms:
64x^3 + 144x^2 + 108x + 27
There is your foiled product.
2009-01-11 12:53 am
= (4x + 3)^3
= (4x)^3 + (3)^3 + 3(4x)(3) [ 4x+ 3 ]
= 64x^3 + 27 + 36x [4x+ 3]
= 64x^3 + 27 + 144x^2 + 108x
= 64x^3 + 144x^2 + 108x + 27
= (4x)^3 + (3)^3 + 3(4x)(3) [ 4x+ 3 ]
= 64x^3 + 27 + 36x [4x+ 3]
= 64x^3 + 27 + 144x^2 + 108x
= 64x^3 + 144x^2 + 108x + 27
2009-01-11 12:49 am
(4x + 3)^3 is already factored. If you want to expand it, then:
(4x + 3)^3
= (4x + 3)(4x + 3)(4x + 3)
= (4x*4x + 3*4x + 4x*3 + 3*3)(4x + 3)
= (16x^2 + 12x + 12x + 9)(4x + 3)
= (16x^2 + 24x + 9)(4x + 3)
= 16x^2*4x + 24x*4x + 9*4x + 16x^2*3 + 24x*3 + 9*3
= 64x^3 + 96x^2 + 36x + 48x^2 + 72x + 27
= 64x^3 + 96x^2 + 48x^2 + 36x + 72x + 27
= 64x^3 + 144x^2 + 108x + 27
(4x + 3)^3
= (4x + 3)(4x + 3)(4x + 3)
= (4x*4x + 3*4x + 4x*3 + 3*3)(4x + 3)
= (16x^2 + 12x + 12x + 9)(4x + 3)
= (16x^2 + 24x + 9)(4x + 3)
= 16x^2*4x + 24x*4x + 9*4x + 16x^2*3 + 24x*3 + 9*3
= 64x^3 + 96x^2 + 36x + 48x^2 + 72x + 27
= 64x^3 + 96x^2 + 48x^2 + 36x + 72x + 27
= 64x^3 + 144x^2 + 108x + 27
2009-01-11 12:43 am
answer is one since we get (4x+3)(4x+3)(4x+3) so 1is common
2009-01-11 12:42 am
This is already factored as much as possible. It can be expanded, which is a little more work, but very doable
First you multiply (4x+3)(4x+3) = 16x^2+24x+9
Next you multiply (16x^2+24x+9)(4x+3) which comes out to 64x^3+144x^2+108x+27
First you multiply (4x+3)(4x+3) = 16x^2+24x+9
Next you multiply (16x^2+24x+9)(4x+3) which comes out to 64x^3+144x^2+108x+27
2009-01-11 12:53 am
To factor this, write it as (4x+3)(4x+3)(4x+3).
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