Asked by BS
Match each polynomial in standard form to its equivalent factored form.
Standard forms:
8x^3+1
2x^4+16x
x^3+8
the equivalent equation that would match with it
(x+2)(x2−2x+4)
The polynomial cannot be factored over the integers using the sum of cubes method.
(2x+16)(4x^2−32x+64)
(x+1)(4x^2−2x+1)
2x(x+2)(x^2−2x+4)
(x+8)(x^2−16x+64)
(2x+1)(4x^2−2x+1)
For equation 1) 8x^3+1 I believe the matching product is (x+8)(x^2−16x+64)
For equation 2) (2x+16)I believe the matching product is (4x^2−32x+64)
For equation 3) x^3+8 I believe the matching product is (x+1)(4x^2−2x+1)
I am not really sure at all I am struggling with this subject
Standard forms:
8x^3+1
2x^4+16x
x^3+8
the equivalent equation that would match with it
(x+2)(x2−2x+4)
The polynomial cannot be factored over the integers using the sum of cubes method.
(2x+16)(4x^2−32x+64)
(x+1)(4x^2−2x+1)
2x(x+2)(x^2−2x+4)
(x+8)(x^2−16x+64)
(2x+1)(4x^2−2x+1)
For equation 1) 8x^3+1 I believe the matching product is (x+8)(x^2−16x+64)
For equation 2) (2x+16)I believe the matching product is (4x^2−32x+64)
For equation 3) x^3+8 I believe the matching product is (x+1)(4x^2−2x+1)
I am not really sure at all I am struggling with this subject
Answers
Answered by
oobleck
8x^3+1 = (2x)^3 + 1^3 = ((2x)+1)((2x)^2 - (2x)(1) + 1^2)
= (2x+1)(4x^2-2x+1)
2x^4+16x = 2x(x^3+1) = 2x(x+1)(x^2-x+1)
x^3+8 = x^3 + 2^3 = ...
= (2x+1)(4x^2-2x+1)
2x^4+16x = 2x(x^3+1) = 2x(x+1)(x^2-x+1)
x^3+8 = x^3 + 2^3 = ...
Answered by
BS
would the third one be (x+2)(x2−2x+4)
Answered by
BS
or The polynomial cannot be factored over the integers using the sum of cubes method.
Answered by
oobleck
your factoring is correct.
sum and difference of cubes can always be factored.
sum and difference of cubes can always be factored.
Answered by
Ace
2x^4+16x = 2x(x^3+1) = 2x(x+1)(x^2-x+1)
Is not a choice I am confused.
Is not a choice I am confused.
Answered by
BS
Yeah I was looking at that and was wondering of I messed up some where
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