136 Easy Algebra Step-by-Step
d. 8 x^3 − 27
Step 1. Observe that the binomial has the form “quantity cubed minus
quantity cubed,” so it is the difference of two cubes. Indicate that
8 x^3 − 27 factors as the product of a binomial and a trinomial. The
binomial has a minus sign between the terms, and the trinomial has
plus signs between the terms.
8 x^3 − 27 = ( − )( ++ )Step 2. Fill in the terms of the binomial. The fi rst term is^382 xx^32 , and
the second term is^327 = 3.
8 x^3 − 27 = ( 2 x− 3 )( ++ )Step 3. Fill in the terms of the trinomial by using the
terms of the binomial, 2x and 3, to obtain the
terms. The fi rst term is (2x)^2 = 4 x^2 , the second
term is 2x⋅ 3 = 6 x, and the third term is 3^2 = 9.
8 x^3 − 27 = ( 2 x− 3 )( 4 x^2 + 6 x+ 9 ) is the factored
form.e. 641 a^3 + 25
Step 1. Observe that the binomial has the form “quantity cubed plus quan-
tity cubed,” so it is the sum of two cubes. Indicate that 64 a^3 + 125
factors as the product of a binomial and a trinomial. The binomial
has a plus sign between the terms, and the trinomial has one minus
sign on the middle term.
64 a^3 + 125 = ( + )( −+ )Step 2. Fill in the terms of the binomial. The fi rst term is^364 aa^344 , and
the second term is^3125 = 5.
64 a^3 + 125 = ( 4 a+ 5 )( −+ )Step 3. Fill in the terms of the trinomial by using the terms of the
binomial, 4a and 5, to obtain the terms. The first term is (4a)^2
= 16 a^2 , the second term is 4a ⋅ 5 = 20 a, and the third term is 5^2
= 25.
64 a^3 + 125 = ( 4 a + 5 )( 16 a^2 − 20 a + 25 )4 x^2 + 6 x + 9 ≠ (2x + 3)^2.
(2x + 3)^2 = 4 x^2 + 12 x +
9, not 4x^2 + 6 x + 9.