Factor.
(4x^2+8r+3)=( )( )
I see the factors of 3, probably 1 and 3
how to make eight? twice 1+3
(2x+3)(2x+1) check that.
Yeah....but second letter is not x.... :-(
oops. Then it has no factors, unless the 3 is 3r^2
ok thanks
To factor the polynomial (4x^2 + 8r + 3), we need to find two expressions that, when multiplied together, equal the original polynomial.
Let's start by looking for two binomials in the form (ax + b) and (cx + d), where a, b, c, and d are constants.
The first term in each binomial will be the product of the coefficients of x^2, which is 4x * 4x = 16x^2.
Next, we need to find the values that will give us the constant term of 3 when multiplied together. Since 3 is a prime number, the only possibilities are 3 and 1, or -3 and -1.
Now, we need to choose the coefficients for x in each binomial. To find these coefficients, we consider the middle term in the original polynomial, which is 8r.
We are looking for two numbers that add up to 8 and multiply to give us the constant term (3).
If we choose the coefficients for the x term as 1 and 3, we can get (1 * 3) + (4x * 2r) = 3 + 8r.
Therefore, the factored form of the original polynomial (4x^2 + 8r + 3) is:
(4x + 1)(x + 3)