Expand the following (4a-3) (4a+3) =

To do this problem, always think of use distributive law twice.

( 4a -3 ) ( 4a+3) = (4 a +3 ) 4a - 3 (4a +3) =

16a

^{2 }+ 12a - 12a - 9 = 16a^{2 }-9Expand the following (4a-3) (4a+3) =

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Woodland Hills, CA

To do this problem, always think of use distributive law twice.

( 4a -3 ) ( 4a+3) = (4 a +3 ) 4a - 3 (4a +3) =

16a^{2 }+ 12a - 12a - 9 = 16a^{2 }-9

Mesa, AZ

When you expand binomials:

First you take the first part of the first expression (4a) times the first part of the second expression, which also happens to be (4a). Your answer to this part of the problem is
**16a**^{2}.

Next, you deal with the middle term. In this case the middle term of the expanded equation is determined by the (-3) in the first binomial and the (4a) in the second binomial. Multiply them together to get the first number that will be combined, and that number is **-12a**. Now combine the 4a from the first binomial with the (+3) in the second binomial to get the second term that will be combined, and that number is
**12a**. Since -12a +12a equals zero, there is no middle term for this answer. Writing a "0" as a space holder is unnecessary. (When combining terms in the expression to get the middle term for the answer, multiply the two outside terms together, multiply the two inside terms together, and add those two answers together to get the middle term of the expansion.)

To get the third term for this expansion, multiply (-3) x (+3) to get -9.

So the answer for this expansion is 16a^{2}-9.

Cary, NC

For this you need to FOIL:

Multiply the first terms (4a)(4a) = 16a^2

Multiply the outer terms (4a)(3) = 12a

Multiply the inner terms (-3)(4a) = -12a

Multiply the last terms (-3)(3) = -9

Now you have:

16a^2 + 12a - 12a - 9

Combine like terms (12a - 12a cancel out) to get:

16a^2 - 9

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## Comments

^{2}+ aX + bX +ab = X^{2 }+( a +b) X + ab^{2}+ ( a + b) X + ab^{2}+ ( a+ b ) X + ab = 0 always factored (X+a) ( X+b) =0_{1}= -a X +b = 0 X_{2}= -b^{2}+ ( a +b ) X + ab = X^{2}+ ( X_{1 +}X_{2}) X + X_{1}X_{2}^{2}+ bX + c =a (X^{2}+b/a X + c/a)_{1 +}X_{2 = }-b/a X_{1}. X_{2}= c/a^{2}+ 8X + 5 = ( 3X +5 ) ( X + 1)_{ X1 = -5/3 }X_{2 = }-1_{ }Where X_{1}+ X_{2}= -5/3 - 1 = -8/3_{1}. X_{2 =}(-5/3) ( -1 ) = 5/3^{2}- a^{2 }( where a+b equals a - a =0)^{2}= ( X + a) ( X + a) = X^{2}+ 2a + a^{2}where a= b , and a +a = 2a