determine whether each trinomial is a perfect square and then factor it.?

Question

I am totally lost with this one

x^2-24x+48

Answer 1

It's not a perfect square.

There are a few quick and easy (yes, easy) telltale signs! I'll tell you those in just a sec after I show you the formula.


For it to be a perfect square, it would fit this formula.

(ax+b)^2 = a^2x^2 + abx + abx + b^2
or = a^2x^2 + 2abx + b^2

Do you see some of the easy ways to tell?

1) a is squared, so if your first number isn't a perfect square, no dice

2) b is squared, so if your last number isn't a perfect square, no dice

3) the middle expression, 2ab, is ALWAYS even

In your case x^2 is ok because 1^2 = 1

but 48 is not a perfect square, so we can toss it out as a perfect square on that alone.

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x^2-24x+48

To factor, we just put the numbers in this form.

(ax + y) (bx + z) format


a x b = 1, so that's a gimme

Then for y and z, it should = 48, right?

Factor 48 and get most likely combos of 4&12 6&8 plus less likely 3 & 16, or 2 & 24

Since a and b are both 1, we can take another shortcut and add our factors to see which will fit our equation

4&12 =16
6&8 = 14
3 & 16 = 19
or 2 & 24 =26

Hmm, none of them = 24, therefore, we're done
x^2-24x+48 is already simplified!

If it were x^2-26x+48, then we'd have (x-2)(x-24)
If it were x^2-19x+48, then we'd have (x-3)(x-16)
If it were x^2-14x+48, then we'd have (x-6)(x-8)
If it were x^2-16x+48, then we'd have (x-12)(x-4)

Does that make sense? See how it works and how just the middle changes? Pretty cool eh?

Answer 2

x^2-24x+48 cannot be a perfect square at the outside as 48 is not a perfect square as to be a perfect square itshould be in the form of (a)^2+/-2(a)(b)+(b)^2

Answer 3

It's not a perfect square. A perfect square would be x^2+14X+49 = (X+7)^2 for example. I would think that the last number in the expression must be able to be square-rooted into a whole number for it to be a square.

Answer 4

(x+24)(x-2)
is's backwords FOIL


yes it is a per fect square