How do you solve # (x+5)(x^2-7x+12)=0#?

Question

Eva Adamson · Accepted Answer

The solutions are #-5#, #3# and #4#.

A multiplications gives zero as a result if and only if at least one of his factors equals zero. So that's what you need to impose. The first factor is quite simple: #x+5# equals zero if and only if #x=-5#. The second factor is a parabola, whose zeroes you may find through the classical formula #{-b \pm \sqrt{b^2-4ac}/2a#, but in simple cases as this, I prefer this simplier one: if the coefficient of #x^2# is one, then you can read your equation like this: #x^2-sx+p=0#, where #s# is the sum of the roots, and #p# is their product. So, you have #-s=-7#, and #p=12#. This means that we are looking for two numbers #a# and #b# such that #a+b=7#, and #a*b=12#. It's easy to see, with barely no calculations, that these numbers are #3# and #4#. So, your equation is solved by three numbers, namely #-5#, #3# and #4#.

Julia Adams · Answer

undefined To solve the equation (x+5)(x^2-7x+12)=0, you first need to set each factor equal to zero and solve for x individually.

Set x + 5 = 0:
x = -5

Now, set x^2 - 7x + 12 = 0 and factor the quadratic equation:
(x - 3)(x - 4) = 0

Set each factor equal to zero:
x - 3 = 0
x = 3

x - 4 = 0
x = 4

So, the solutions to the equation (x+5)(x^2-7x+12)=0 are x = -5, x = 3, and x = 4.