How to factor trimonials

Question

What are the stops to solve this by factoring 13z² + 81z − 7

Answer 1

As a quick check, if we are to have any hope of factoring a quadratic equation (of the form  ax²+bx+c = 0) into linear factors with rational coeffecients we must have that the discriminant of the quadratic equation (which is defined to be:  b² - 4ac) is a perfect square. Since a = 13, b = 81, and c = -7 here, the discriminant is: 

 

81² - 4 * (13)* (-7) = 6925

 

You can quickly see that 6925 is not a perfect square (since √6925  is not an integer). Thus we cannot factor this expression into linear factors with rational coefficients no matter how we try. 

Answer 2

Two of your three coefficients are prime numbers, so you cannot factor them out. The closest you can come (to SIMPLIFY the expression, instead of just changing it) is to pull out the z: then you'd have z(13z+81) -7

Answer 3

This is an incredibly ugly problem.

first let us assume that we look at every quadratic equation in the form a*x^2 + b*x + c so in this case a is 13, b is 81 and c is -7.

Normally I would suggest looking at all the possible factorings of a and c, but in this case the only factoring of 13 is 13 and 1 and the only factorings of -7 are (1 and -7) or (-1 and 7) which for our purposes are pretty similar.

The only equations that would be kind to you when factoring that started with 13 and ended with -7 would be:

13z^2 + 90z - 7,

13z^2 - 90z - 7,

13z^2 + 6z - 7, or

13z^2 - 6z - 7

So normally if you split up the number a into factors (f1 and f2)

and the number c into the factors (g1 and g2) then the answer will eventually look like this:

((f1 * z) + g1) * ((f2 * z) + g2) which means the middle number b will be equal to ((f1 * g2) + (f2 * g1))

Which is how I found the four equations above.

That is not to say that this equation is unsolvable, but in order to solve it you would either use completing the square (which would be VERY ugly as well)

or you would use the quadratic equation where using the same lettering system you could find the zeroes of the equation:

z = ((-b (+ or -) sqrt(b^2 - 4ac)) / 2a)

(if you look up the "quadratic equation" you will find a much nicer looking version of what I just wrote).

Then if you make z1 = -b + blahdy blah

and z2 = -b - blahdy blah

(notice z1 uses the plus, and z2 uses the minus but the rest of the equation will stay the same, I just didn't want to type it out) then you could factor it by doing this:

(z - z1) * (z - z2)

but this will end up with square roots and is just plain ugly.

I would double check your numbers. Please let me know if you have any further questions, this is a complicated one to answer through text.