3x² + 8x + 4
First, i divide the equation into two parenthesis, so that the first parts of both multiply to make the first term, 3x². 3x * x = 3x²
(3x + )(x + )
Then I find two numbers that multiply to make 4, which are 1 and 4, or 2 and 2.
Our options are:
(3x + 1)(x + 4)
(3x + 4)(x + 1)
(3x + 2)(3x +2)
To figure out which one to use, I'm just going to FOIL them all.
(3x + 1)(x + 4) = 3x² + 1x + 12x + 4 = 3x² + 13x + 4
(3x + 4)(x + 1) = 3x² + 4x + 3x + 4 = 3x² + 7x + 4
(3x + 2)(x +2) = 3x² + 2x + 6x + 4 = 3x² + 8x + 4
The factored form is:
(3x + 2)(x + 2)
To solve for the roots ( where the graph crosses the x axis, where y = 0) we set the equation equal to 0:
(3x + 2)(x + 2) = 0
The zero product property says that anything times 0 is 0, so we set each individual part equal to 0 and solve for the two roots.
3x + 2 = 0
3x = -2
x = -2/3
x + 2 = 0
x = -2
Answer:
500
Step-by-step explanation:
I assume each number is from 0 to 9. Also there are 5 letters.
10 * 5 * 10 = 500
Answer:
Step-by-step explanation:
The algebraic expression states that
"The class we divided into four equal groups."
We have our expression right at the word of "divided", therefore the answer would be a division problem.
w over 4 fits this because it fits the expression, w - class and 4 - amount of groups.
4 + w - Would be the class added four groups
4w - The class multiplies four more groups
4 - w - Four groups was taken from the class
Answer:
the answer to your question is bbbbb
Step-by-step explanation:
B = {a, b, c, d}
C = {0, a, 2, b}
B ∪ C = {a, b, c, d, 0, 2}
<h3>Answer: E)</h3>
Everything from the set B and everything from the set C give to one set. Duplicated elements are written only once.
