Factor Polynomials of Quadratic Form: (4j - 2)^2 - (2 + 4j)^2

[VixxenBlack]

Please help. the question asks to factor each polynomial in quadratic form.

a) 6(x2 - 4x + 4)^2 + (x2 - 4x + 4) - 1


b)(4j - 2)^2 - (2 + 4j)^2

 

[Deleted member 4993]

Please help. the question asks to factor each polynomial in quadratic form.

a) 6(x2 - 4x + 4)^2 + (x2 - 4x + 4) - 1


b)(4j - 2)^2 - (2 + 4j)^2

What are your thoughts?

Please share your work with us ...even if you know it is wrong.

If you are stuck at the beginning tell us and we'll start with the definitions.

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[Steven G]

Please help. the question asks to factor each polynomial in quadratic form.

a) 6(x2 - 4x + 4)^2 + (x2 - 4x + 4) - 1


b)(4j - 2)^2 - (2 + 4j)^2

a) Let u = something that should be obvious and then get a quadratic in terms of u. Have you been attending class? Have you been listening and taking notes?
b) have you ever heard of difference of squares.

i truly believe that you do not know where to start so I am giving you hints.

 

[stapel]

Please help. the question asks to factor each polynomial in quadratic form.

a) 6(x2 - 4x + 4)^2 + (x2 - 4x + 4) - 1


b)(4j - 2)^2 - (2 + 4j)^2

They've taught you how to factor regular quadratics. They've given you the hint that they polynomials are "in quadratic form", so you know that you can rewrite them as quadratics. What have you done with this information? For instance, you started (a) with the obvious substitution:

. . . . .6Z² + X - 1

You factored in the usual way (here), plugged x² - 4x + 4 back in for Z, simplified, and... then what? Same questions for (b).

Please be complete. Thank you!