How do you factor #y=x^3+8x^2+17x+10 #?

The rational root theorem states that in a polynomial #ƒ(x) = qx^n + mx^(n - 1) + ... + px^0#, the possible roots will be at #p/q#.

Therefore, we can state that the possible roots are at (factors of 10)/(factors of 1). Listing these, we get:

#(+- 1, +-2, +-5, +-10)/(+-1)#

Simplifying further:

#+-1, +-2, +-5, +-10#

Here comes the part that can be long. The remainder theorem states that for any polynomial function #p(x)# being divided by #x - a)#, the remainder is given by evaluating #p(a)#. A factor, when evaluated, will give a final result, or a remainder, of 0, since by definition a factor is a number that evenly divides another.

When I do this, I tend to go from positive to negative, from smallest to largest. From experience, it is relatively rare for you not to find a factor by #(x + 2)#. Following my technique, we will start with #(x - 1)# and then proceed with #(x + 1)#. Inspect the following proofs:

#y = (1)^3 + 8(1)^2 + 17(1) + 10#

#y = 1 + 8 + 17 + 10#

#y = 36#, therefore, #(x - 1)# is not a factor of this polynomial.

#y = (-1)^3 + 8(-1)^2 + 17(-1) + 10#

#y = -1 + 8 - 17 + 10#

#y = 0#

Therefore, #(x + 1)# is a factor of this polynomial.

We must now divide #(x^2 + 8x^2 + 17x + 10)# by #(x + 1)# to see what is left over.

The following image shows the long division (you could have also used synthetic division).

As you can see, the quotient is #x^2 + 7x + 10#

This can easily be factored by finding two numbers in #y = ax^2 + bx + c, a = 1# that multiply to c and that add to b. Two numbers that do this are #5 and 2#

Therefore,

#y = x^3 + 8x^2 + 17x + 10 = (x + 1)(x + 5)(x + 2)#

or

#y = (x + 1)(x + 5)(x + 2)#

Hopefully this helps!