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Is it possible to factor #y=9x^2-48x+64#? If so, what are the factors?

Answer 1

Yes

If you are having a problem factorising a quadratic by inspection then use the formula #x=(-b+-sqrt(b^2-4ac))/(2a)#

In this example the discriminant #b^2-4ac#=0 So the factors are the same #x=48/(2*9)=8/3# So the two factors are both #3x-8#

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Answer 2

Genesis Allen

Yes, it is possible to factor the quadratic equation (y = 9x^2 - 48x + 64). To factor it, we can use the quadratic formula or factoring by grouping.

First, let's check if the expression can be factored by grouping:

[y = 9x^2 - 48x + 64]

We can rewrite the expression as:

[y = 9x^2 - 36x - 12x + 64]

Now, we can factor by grouping:

[y = (9x^2 - 36x) + (-12x + 64)]

Taking out the common factors from each group:

[y = 9x(x - 4) - 4(3x - 16)]

Now, we can see that both terms have a common factor of (3x - 16), so we can factor it out:

[y = (9x - 4)(x - 4)]

So, the factored form of the quadratic equation (y = 9x^2 - 48x + 64) is (y = (9x - 4)(x - 4)).

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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.