4x^2 + 4x + 1
Can it be factored? If so, how?
2007-08-31 19:02:29 · 10 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
(2x + 1) (2x + 1) are the factors
2007-09-01 08:45:05 · answer #1 · answered by Como 7 · 2⤊ 0⤋
Yes, it can be factored.
You would have to know something called algebraic identities. This question is an algebraic identity in itself. There are 3 basic algebraic identities :
1. (a + b)(a + b) = a^2 + 2ab + b^2
2. (a - b)(a - b) = a^2 - 2ab + b^2
3. (a - b)(a + b) = a^2 - b^2
In this question we use the first algebraic identity.
We substitute a and b into the question.
a^2 + 2ab + b^2 = 4x^2 + 4x + 1,
= (2x)^2 + 2(2x) + 1^2
So now we know that a = 2x and b = 1.
Substitute a = 2x and b = 1 into (a + b)(a + b)
We get (2x + 1)(2x +1).
The entire equation shows :
(2x + 1)(2x + 1) = 4x^2 + 4x + 1
I hope this helps.
2007-08-31 22:40:07 · answer #2 · answered by Answerer 2 · 0⤊ 1⤋
Yes this quadratic equation is solvable by factoring.
In fact:
4x^2 + 4x + 1 = (2x+1)(2x+1)
2007-08-31 19:08:12 · answer #3 · answered by Anonymous · 1⤊ 0⤋
this is a perfect square trinomial, so yes it can be solved by factoring. as those before me have said, the answer is (2x+1)^2.
the way to recognize a perfect square is simple. if both the constant and the coefficient of x^2 are perfect squares (4 and 1 in this case) and twice the product of their square roots equals the coefficient of x (2*2*1=4), then you have a perfect square trinomial.
9x^2-12x+4 is another one (3*2*2=12). it would factor as (3x-2)^2. note that the sign in the factors always matches the sign on the coefficient of x.
hope this helps
2007-08-31 19:19:52 · answer #4 · answered by Ryan J 2 · 0⤊ 0⤋
No. in case you could remodel the equation into the commonplace style. Ax^2 + Bx + C = 0; that's a similar situation as rational. in case you could’t get it interior the commonplace style then the equation ought to be solved by the two of two concepts: a million. polishing off the sq.. placed into commonplace style ax^2 + bx =c. A !=a million or not equivalent to a million then divide the equation by a… The consistent fee (b/2a)^2 is to be further to the two factors. Then write the LHS of the equation as a pleasing sq.. Take sq. roots of the two factors ….. resolve for x. 2. making use of the quadratic formula. X = -b +/- sq. root (b^2 - 4ac)/2a
2016-12-16 08:20:19 · answer #5 · answered by ? 4 · 0⤊ 0⤋
= 4x^2 + 4x + 1
= (2x + 1) (2x + 1)
2007-09-04 18:58:37 · answer #6 · answered by Jun Agruda 7 · 3⤊ 0⤋
4x^2 + 4x + 1
4x^2 +2x +2x +1
2x(2x+1) + (2x+1)
(2x+1)(2x+1) Two factors
2007-08-31 19:10:13 · answer #7 · answered by Indian Primrose 6 · 1⤊ 0⤋
(2x + 1)^2
2007-08-31 19:09:52 · answer #8 · answered by ǝɔnɐs ǝɯosǝʍɐ Lazarus'd- DEI 6 · 0⤊ 0⤋
(2x + 1)²
2007-08-31 19:08:17 · answer #9 · answered by chasrmck 6 · 0⤊ 0⤋
= (2x + 1) (2x + 1)
2007-08-31 19:08:43 · answer #10 · answered by Hk 4 · 0⤊ 0⤋