How do you factor y=x^2 + 35x + 36 ?

Dec 27, 2015

You can not factor that trinomial, unfortunately.

Explanation:

You can always see if a quadratic function like the one above is factorable by using the discriminant of the quadratic formula:

${b}^{2}$ - 4ac
${35}^{2}$ - 4(1)(36)
1120 - 144
976

Since the final result, 976, is not a perfect square, this trinomial cannot be factored into even factors.

Hopefully you understand now.

Dec 27, 2015

This quadratic can only be factored using irrational coefficients:

${x}^{2} + 35 x + 36$

$= \left(x + \frac{35}{2} - \frac{\sqrt{1081}}{2}\right) \left(x + \frac{35}{2} + \frac{\sqrt{1081}}{2}\right)$

Explanation:

The difference of squares identity can be written:

${a}^{2} - {b}^{2} = \left(a - b\right) \left(a + b\right)$

So:

${x}^{2} + 35 x + 36$

$= {x}^{2} + 35 x + {\left(\frac{35}{2}\right)}^{2} - {\left(\frac{35}{2}\right)}^{2} + 36$

$= {\left(x + \frac{35}{2}\right)}^{2} - \frac{1225}{4} + \frac{144}{4}$

$= {\left(x + \frac{35}{2}\right)}^{2} - \frac{1081}{4}$

$= {\left(x + \frac{35}{2}\right)}^{2} - {\left(\frac{\sqrt{1081}}{2}\right)}^{2}$

$= \left(\left(x + \frac{35}{2}\right) - \frac{\sqrt{1081}}{2}\right) \left(\left(x + \frac{35}{2}\right) + \frac{\sqrt{1081}}{2}\right)$

$= \left(x + \frac{35}{2} - \frac{\sqrt{1081}}{2}\right) \left(x + \frac{35}{2} + \frac{\sqrt{1081}}{2}\right)$

We cannot simplify $\sqrt{1081}$ further since $1081 = 23 \cdot 47$ has no square factors.