How do you find the product #(x+3)(x+3)#?

Answer 1

Genesis Allen

See a solution process below:

This is a special form of quadratic where:

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

Substituting #x# for #a# and #3# for #b# gives:

#(x + 3)(x + 3) => (x + 3)^2 => x^2 + 2x3 + 3^2 =>#

#x^2 + 6x + 9#

If you want to multiply out the long way you can use this process. To multiply these two terms you multiply each individual term in the left parenthesis by each individual term in the right parenthesis.

#(color(red)(x) + color(red)(3))(color(blue)(x) + color(blue)(3))# becomes:

#(color(red)(x) xx color(blue)(x)) + (color(red)(x) xx color(blue)(3)) + (color(red)(3) xx color(blue)(x)) + (color(red)(3) xx color(blue)(3))#

#x^2 + 3x + 3x + 9#

We can now combine like terms:

#x^2 + (3 + 3)x + 9#

#x^2 + 6x + 9#

The same result as above.

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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.