Is it possible to factor y=2x^2 + 8x+4y=2x2+8x+4? If so, what are the factors?
2 Answers
First you can take out the factor
Explanation:
This cannot be factored any further without using radical expressions.
2x^2+8x+4 = 2(x^2+4x+2)2x2+8x+4=2(x2+4x+2)
= 2(x+2+sqrt(2))(x+2-sqrt(2))=2(x+2+√2)(x+2−√2)
Explanation:
Given
y = 2(x^2+4x+2)y=2(x2+4x+2)
The quadratic factor is in the form
x = (-b+-sqrt(b^2-4ac))/(2a) = (-4+-sqrt((-4)^2-(4xx1xx2)))/(2xx1)x=−b±√b2−4ac2a=−4±√(−4)2−(4×1×2)2×1
=(-4+-sqrt(8))/2 = (-4+-sqrt(2^2*2))/2 = (-4+-2sqrt(2))/2 = -2+-sqrt(2)=−4±√82=−4±√22⋅22=−4±2√22=−2±√2
These zeros correspond to factors
Another way of finding this is by completing the square, then using the difference of squares identity:
a^2-b^2 = (a-b)(a+b)a2−b2=(a−b)(a+b)
with
y = 2x^2+8x+4y=2x2+8x+4
=2(x^2+4x+2)=2(x2+4x+2)
=2(x^2+4x+4-2)=2(x2+4x+4−2)
=2((x+2)^2-2)=2((x+2)2−2)
=2((x+2)^2-(sqrt(2))^2)=2((x+2)2−(√2)2)
=2((x+2)-sqrt(2))((x+2)+sqrt(2))=2((x+2)−√2)((x+2)+√2)
=2(x+2-sqrt(2))(x+2+sqrt(2))=2(x+2−√2)(x+2+√2)
