Question

hi :)
1(a^8) + 4(a^4)b +4(b^2)-1(c^4)
factor this to products of two binomial

Answer 1

\( 1(a^8) + 4(a^4)b +4(b^2)-1(c^4) \)

\( = a^8 + 4a^4b +4b^2-c^4 \)

\(= a^8 + (4b)a^4 + (2b + c^2)(2b - c^2) \)

The expression above is of the form \(x^2 + bx + c\)
where \(a = 1\)
\(b = 4b\), and
\(c = (2b + c^2)(2b - c^2)\)

To factor it, we need to factors of c that add to b.
We have two factors of c already: \(2b + c^2\) and \(2b - c^2\)
We need the factors to add to the middle coefficient which is 4b.
\(2b + c^2 + 2b - c^2= 4b\)
It works, so we can factor:

\(= a^8 + (4b)a^4 + (2b + c^2)(2b - c^2) \)

\(= (a^4 + 2b + c^2)(a^4 + 2b - c^2)\)

This was factored into two trinomials, not two binomials.

Answer 2

\( 1(a^8) + 4(a^4)b +4(b^2)-1(c^4) \)

\( = a^8 + 4a^4b +4b^2-c^4 \)

\(= a^8 + (4b)a^4 + (2b + c^2)(2b - c^2) \)

The expression above is of the form \(x^2 + bx + c\)
where \(a = 1\)
\(b = 4b\), and
\(c = (2b + c^2)(2b - c^2)\)

To factor it, we need to factors of c that add to b.
We have two factors of c already: \(2b + c^2\) and \(2b - c^2\)
We need the factors to add to the middle coefficient which is 4b.
\(2b + c^2 + 2b - c^2= 4b\)
It works, so we can factor:

\(= a^8 + (4b)a^4 + (2b + c^2)(2b - c^2) \)

\(= (a^4 + 2b + c^2)(a^4 + 2b - c^2)\)

This was factored into two trinomials, not two binomials.