Both terms are divisible by 4x^44x4 so we can separate out that factor first...
8x^4-28x^6 = 4x^4(2-7x^2)8x4−28x6=4x4(2−7x2)
Next, if we allow irrational coefficients, 2-7x^22−7x2 can be treated as a difference of squares, so we can use the difference of squares identity:
a^2-b^2 = (a-b)(a+b)a2−b2=(a−b)(a+b)
with a=sqrt(2)a=√2 and b=sqrt(7)xb=√7x
So:
2-7x^2 = (sqrt(2))^2 - (sqrt(7)x)^2 = (sqrt(2)-sqrt(7)x)(sqrt(2)+sqrt(7)x)2−7x2=(√2)2−(√7x)2=(√2−√7x)(√2+√7x)
Putting it all together:
8x^4-28x^6 = 4x^4(2-7x^2) = 4x^4(sqrt(2)-sqrt(7)x)(sqrt(2)+sqrt(7)x)8x4−28x6=4x4(2−7x2)=4x4(√2−√7x)(√2+√7x)
