factors of 3x^2+4x+1 explain ,thanx?

Answer 1

product of constant and coefficient of x^2 = 1*3 = 3

split 4 (coefficient of x) into 2 terms so that product is 3 1 and 3

so 3x^2 + 4x + 1
= 3x^2+ 3x + x + 1
= 3x(x+1) + (x+1)
= (x+1)(3x+1)

Answer 2

When looking for factors, it's good to start by writing it in ax^2 + bx + c form, as this is already written.

Look for anything that looks the same in all of the pieces of your polynomial, and bring that out. In this case there is nothing to bring out, but one example where it is very good to do would be x^2 + 2x, which equals x(x+2).

Since this is a second degree polynomial (2 is the largest exponent), you know you'll have 2 factors. so draw out two sets of parentheses:
(_____)(_____)

But since it is a second degree polynomial, you know that it has to be some ax times some other bx for some real numbers a and b. so then you may write (ax___)(bx___). but it can't be just (ax)(bx) for obvious reasons. So look for two numbers so that they equal 1. Please note that mathematicians are inherently lazy, and you should always look for the easiest way to multiply for something. So, for example 1=1 times 1. So try (ax+1)(bx+1). Now all we have to do is find a number so that, after FOILing (a way multiplying binomials, if it isn't what your teacher calls it) (ax+1)(bx+1)=(3x^2 + 4x + 1).

but (ax+1)(bx+1) equals
(a)(b)x^2 + (1)(a)x +(1)(b)x + (1)(1)
=abx^2 + ax + bx + 1
=abx^2 + (a+b)x +1
+3x^2 + 4x + 1

So you need ab=3 and a+b=4.

In this case, there is an easy solution, a=3 and b=1.
However, in some cases you won't be as lucky to have c=1 (with the c from the polynomial form on the first line). For example, if c=4, you may have to try 1x4 and 2x2 both before you find an a and b (from the latter equations) that make any sense.

Now for the most important part, check your answer.
Always, Always, Always check your answer.

Answer 3

This is what i would do, first multiply the A and the C, in this case you would get 3 correct? Then you would find two numbers that multiply into 3 but add up to the B term (in this case is four).So the numbers that multiply to 3 but add up to 4 is (3,and 1) because 3times 1 is 3 and 3+1 is four. At the end you would put the A term plus the two numbers which is 3 and 1,so it would look like this

(3x+1)(3x+3)

but ur not done yet, see how the second one is (3x+3)?
NOW DIVEIVE 3 so u would get (x+1) so ur end solution is

(3x+1)(x+1) TO check, foil it out

Answer 4

3x²+4x+1

1.) Set up your FOIL parenthesis with what you know for sure
- (3x _ _ ) (x _ _)
2.) Since the number without a variable, 1, is positive the signs are either going to be +,+ or -,-, but since 4x is positive we know that the signs are going to be +'s
- (3x+ _)(x + _)
3.) We know that the product of the remaining unkowns has to equal one, so we'll "guess and check" with both unknowns being 1
- (3x+1)(x+1)
4.)FOIL and check
F -- 3x * x = x²
O -- 3x * 1 = 3x
I -- 1 * x = x
L -- 1 * 1 = 1

So we have x²+3x+x+1 which equals x² + 4x + 1, so it checks

5.)Now we set (3x+1) and (x+1) equal to zero and solve.
• 3x + 1 = 0 -> 3x = -1 -> x + -1/3
• x + 1 = 0 -> x = -1

So x = -1/3 and x = -1

Answer 5

You want to factor 3x²+4x+1 into something that looks like (ax+b)*(cx+d). Thinking about how binomials are multiplied, (ax+b)*(cx+d) = acx² + (ad+bc)x + bd. So (in your problem) you need to find factors of 3 and 1 (which is no big deal since they're both prime ☺) Then about the only way that they *can* factor is (3x+1)*(x+1).

These are a *lot* more fun if it's something like
120x²+300x-720 ☺


Doug

EDIT: Why is everybody looking for the roots of a quadratic? ☺

Answer 6

3x^2 + 4x + 1 = 0
1/3 (3x + 3) (3x + 1) = 0
(x + 1) (3x + 1) = 0
- x + 1 = 0 ===> x = -1
- 3x + 1 = 0 ===> x = -1/3

-1 and -1/3

Answer 7

3x^2 + 4x + 1 =
(9x^2 + 12x + 3)/3 =
(9x^2 + 12x + 4 - 1)/3 =
[(3x + 2)^2 - 1]/3 =
(3x + 2 + 1)(3x + 2 - 1)/3 =
(3x + 3)(3x + 1)/3 =
(x + 1)(3x + 1)

Which can much easier be arrived at by inspection.

Answer 8

3x^2+4x+1=3x^2+3x+x+1=3x(x+1)+1(x+1)=(x+1)(3x+1)Ans.