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How do you factor #20x^2 - 5#?

Answer 1

Eli Assouline

#20x^2-5=5(4x^(2)-1)# #=# #5(2x-1)(2x+1)#

So we have #20x^2-5=5(4x^(2)-1)#; dividing thru by #5#.

#5(4x^(2)-1)#

And #4x^2-1# is the difference between 2 squares, #2x# and #1#.

#(4x^(2)-1)# #=# #(2x-1)(2x+1)#

Putting it all together:

#20x^2-5=5(2x-1)(2x+1)#

So there are roots at #+-1/2#.

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Answer 2

Jace Anderson

To factor the expression 20x^2 - 5, you can first look for common factors. In this case, there are no common factors. Then, you can use the difference of squares formula, which states that a^2 - b^2 = (a + b)(a - b). Applying this formula, we get: 20x^2 - 5 = (4x)^2 - (√5)^2 = (4x + √5)(4x - √5). Therefore, the factored form of 20x^2 - 5 is (4x + √5)(4x - √5).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.