How do you factor and solve each polynomial completely #y=x^4-14x^2+40#?

1 Answer
George C.
Jan 8, 2017

#x^4-14x^2+40 = (x-sqrt(10))(x+sqrt(10))(x-2)(x+2)#

with zeros: #+-sqrt(10)# and #+-2#

Explanation:

The difference of squares identity can be written:

#a^2-b^2 = (a-b)(a+b)#

We use this a couple of times below.

Note that #10+4 = 14# and #10*4 = 40#

So we find:

#x^4-14x^2+40 = (x^2-10)(x^2-4)#

#color(white)(x^4-14x^2+40) = (x^2-(sqrt(10))^2)(x^2-2^2)#

#color(white)(x^4-14x^2+40) = (x-sqrt(10))(x+sqrt(10))(x-2)(x+2)#

So this quartic has zeros:

#x = +-sqrt(10)" "# and #" "x = +-2#

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