Is it possible to factor $y = 2 x^{2} + 13 x + 6$ $y=2x^2 + 13x + 6$ ? If so, what are the factors?

Question

Is it possible to factor $y = 2 x^{2} + 13 x + 6$ $y=2x^2 + 13x + 6$ ? If so, what are the factors?

1 Answer

Impact of this question

3365 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-01-22T12:28:01

$2 x^{2} + 13 x + 6 = 2 (x + \frac{1}{2}) (x + 6)$ $2x^2 + 13x + 6 = 2 (x + 1/2)(x+6)$

Explanation:

If it is possible to factor

$y = 2 x^{2} + 13 x + 6$ $y = 2x^2 + 13x + 6$ ,

then your equation can be written as

$y = a (x + r) (x + s)$ $y = a (x + r)(x + s)$

Here, $a$ $a$ is the coefficient of the $x^{2}$ $x^2$ term, so $a = 2$ $a= 2$ .

$y = 2 (x^{2} + \frac{13}{2} x + 3)$ $y = 2 (x^2 + 13/2 x + 3)$

$= 2 (x + r) (x + s)$ $= 2 (x + r)(x+s)$

One way to find such numbers $r$ $r$ and $s$ $s$ is computing the roots (zeros) of the polynomial.

To do so, you need to set $x^{2} + \frac{13}{2} x + 3 = 0$ $x^2 + 13/2x + 3 = 0$ and search for possible solutions.

This can be done e.g. with a quadratic formula. However, let me show you one of my favorite methods to find solutions of a quadratic equation: completion of the circle.

If you are not interested and would rather do it with the quadratic formula, you can skip the next part!

======================

$x^{2} + \frac{13}{2} x + 3 = 0$ $x^2 + 13/2 x + 3 = 0$

... compute $- 3$ $-3$ on both sides of the equation...

$x^{2} + \frac{13}{2} x = - 3$ $x^2 + 13/2x = -3$

Now, on the left side, we would like to have something like $a^{2} + 2 a b + b^{2}$ $a^2 + 2ab + b^2$ so that we are able to apply the formula

$a^{2} + 2 a b + b^{2} = {(a + b)}^{2}$ $a^2 + 2ab + b^2 = (a+b)^2$

We already have $a^{2} = x^{2}$ $a^2 = x^2$ , so our $a = x$ $a = x$ , and we also have $2 a b = \frac{13}{2} x$ $2ab = 13/2 x$ . Thus, we can conclude that $b = \frac{13}{4}$ $b = 13/4$ .

Now, to create an $a^{2} + 2 a b + b^{2}$ $a^2 + 2ab + b^2$ expression on the left side, we need to add our $b^{2}$ $b^2$ term, namely ${(\frac{13}{4})}^{2}$ $(13/4)^2$ . However, since we don't want to jeopardize the equality, we need to add ${(\frac{13}{4})}^{2}$ $(13/4)^2$ on the right side of the equation as well:

$x^{2} + \frac{13}{2} x + {(\frac{13}{4})}^{2} = - 3 + {(\frac{13}{4})}^{2}$ $x^2 + 13/2x + (13/4)^2= -3 + (13/4)^2$

... apply $a^{2} + 2 a b + b^{2} = {(a + b)}^{2}$ $a^2 + 2ab + b^2 = (a+b)^2$ on the left side and calculate the right side...

${(x + \frac{13}{4})}^{2} = \frac{121}{16}$ $(x +13/4)^2 = 121/16$

Now, you can draw the root on the left side, but be careful: when doing so, you are creating two solutions since e.g. for $x^{2} = 25$ $x^2 = 25$ , both $x = 5$ $x = 5$ and $x = - 5$ $x = -5$ are solutions.

Thus, we get

$x + \frac{13}{4} = \sqrt{\frac{121}{16}} or x + \frac{13}{4} = - \sqrt{\frac{121}{16}}$ $x + 13/4 = sqrt(121/16) " or " x + 13/4 = - sqrt(121/16)$

$x = - \frac{1}{2} or x = 6$ $x = - 1/2 " or " x = 6$

======================

Thus, the equation

$2 x^{2} + 13 x + 6 = 0$ $2x^2 + 13x + 6 = 0$

$2 (x^{2} + \frac{13}{2} x + 3) = 0$ $2 (x^2 + 13/2 x + 3) = 0$

has two solutions:

$x = - \frac{1}{2} or x = 6$ $x = - 1/2 " or " x = 6$

Thus, you can factorize using negative values of the two solutions:

$2 x^{2} + 13 x + 6 = 2 (x - (- \frac{1}{2})) (x - 6) = 2 (x + \frac{1}{2}) (x + 6)$ $2x^2 + 13x + 6 = 2 (x - ( - 1/2) )(x - 6) = 2 (x + 1/2)(x+6)$

Science

Math

Humanities

... and beyond

Is it possible to factor $y = 2 x^{2} + 13 x + 6$ $y=2x^2 + 13x + 6$ ? If so, what are the factors?

1 Answer

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

Is it possible to factor y=2x^2 + 13x + 6 y=2x2+13x+6y=2x^2 + 13x + 6 ? If so, what are the factors?

1 Answer

Explanation:

Related questions

Impact of this question

Is it possible to factor $y = 2 x^{2} + 13 x + 6$ $y=2x^2 + 13x + 6$ ? If so, what are the factors?