How do you factor 3x ^ { 4} + 12x ^ { 3} + 4x + 243x4+12x3+4x+24?
1 Answer
A few thoughts...
Explanation:
The quartic as it stands has no rational factors.
If there is a typo in the question and one of the terms is in error, then it may be possible to factor by grouping.
For example:
3x^4+12x^3+4x+color(red)(16) = 3x^3(x+4)+4(x+4) = (3x^3+4)(x+4)3x4+12x3+4x+16=3x3(x+4)+4(x+4)=(3x3+4)(x+4)
3x^4+12x^3+color(red)(6)x+24 = 3x^3(x+4)+6(x+4) = 3(x^3+2)(x+4)3x4+12x3+6x+24=3x3(x+4)+6(x+4)=3(x3+2)(x+4)
3x^4+color(red)(18)x^3+4x+24 = 3x^3(x+6)+4(x+6) = (3x^3+4)(x+6)3x4+18x3+4x+24=3x3(x+6)+4(x+6)=(3x3+4)(x+6)
color(red)(2)x^4+12x^3+4x+24 = 2x^3(x+6)+4(x+6) = 2(x^3+2)(x+6)2x4+12x3+4x+24=2x3(x+6)+4(x+6)=2(x3+2)(x+6)
It is possible to factor each of the above possibilities further using irrational coefficients.
For example:
a^3+b^3 = (a+b)(a^2-ab+b^2)a3+b3=(a+b)(a2−ab+b2)
So:
x^3+2 = x^3+(root(3)(2))^3 = (x+root(3)(2))(x^2-root(3)(2)x+root(3)(4))x3+2=x3+(3√2)3=(x+3√2)(x2−3√2x+3√4)
Alternatively, if the question is correct as given, then it is possible to factorise, but the answer gets very messy and complicated. I may write up a separate answer to sketch the process.