How do you factor the binomial $4x^2 + 15x + 10$ ?

Question

2111 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-06-14T19:02:54

Use quadratic formula to find zeros of $4x^2+15x+10$ and hence factors:

$4x^2+15x+10 = (2x+(15+sqrt(65))/4)(2x+(15-sqrt(65))/4)$

$f(x) = 4x^2+15x+10$ is of the form $ax^2+bx+c$ with $a=4$ , $b=15$ and $c=10$

The discriminant is given by the formula:

$Delta = b^2-4ac = 15^2-(4xx4xx10) = 225 - 160 = 65$

This is positive but not a perfect square, so the roots of $f(x) = 0$ and coefficients of the linear factors are real but irrational.

$f(x)=0$ has roots given by the quadratic formula

$x = (-b +- sqrt(Delta))/(2a) = (-15+-sqrt(65))/8$

Since the leading term we are aiming for is $4x^2$ , multiply this through by $2$ to get:

$2x = (-15+-sqrt(65))/4$

Hence $f(x)$ may be factored as:

$(2x+(15+sqrt(65))/4)(2x+(15-sqrt(65))/4)$

How do you factor the binomial 4x^2 + 15x + 104x^2 + 15x + 10?