Factoring a polynomial, and in particular a quadratic polynomial, is analogous to factoring a positive integer. For instance, we know that 65=5*13, where 5 and 13 are each factors of 65. For positive integers, we have the fundamental theorem of arithmetic, which basically says that every positive integer greater than...

## See

This Answer NowStart your **48-hour free trial** to unlock this answer and thousands more. Enjoy eNotes ad-free and cancel anytime.

Already a member? Log in here.

Factoring a polynomial, and in particular a quadratic polynomial, is analogous to factoring a positive integer. For instance, we know that 65=5*13, where 5 and 13 are each factors of 65. For positive integers, we have the fundamental theorem of arithmetic, which basically says that every positive integer greater than 1 factors uniquely, up to order, in the prime numbers. Thus, the factorization of 65 is 5*13, as 5 and 13 are prime. 12=2*2*3 is the unique factorization of 12.

For polynomials, we have the fundamental theorem of algebra. One consequence of the fundamental theorem is that every polynomial with real coefficients factors uniquely into linear factors over the complex numbers.

Instructions for quadratics are usually of three types. (1) Evaluate: find the value of the function for some input (often used for graphing). (2) Solving: where the polynomial is set equal to zero and we are asked to find the zeros or roots. (3) Factoring: writing the polynomial as the product of linear factors. In the case of factoring, we are often asked to factor with some restrictions. `x^2+1` is not factorable over the reals (it is said to be irreducible) but factors as (x+i)(x-i) over the complex numbers. `x^2-2` is irreducible over the rationals but factors as `(x+sqrt(2))(x-sqrt(2))` over the reals.

There are many methods to factor a quadratic.

(1) The AC method: If we have `Ax^2+Bx+C`, then we attempt to find p,q such that pq=AC and p+q=B. This only works well if the quadratic is factorable over the rationals.

`x^2+4x-21=(x+7)(x-3)` Here, we have AC=-21 and B=4, so we look for p,q such that pq=-21 and p+q=4, if p=7 and q=-3. Then we can rewrite the original quadratic as `x^2+7x-3x-21=x(x+7)-3(x+7)=(x+7)(x-3)`

** After finding p,q we use factoring by grouping.**

(2) We can use guess and check; Suppose we have `2x^2-7x-15` We know it has to factor as (2x+?)(x-?) or (2x-?)(x+?). When multiplying the binomials, the first term must be `2x^2`, and the last term is negative, so we must multiply a positive and negative number. After that, we try factors of 15 in the unknown slots until we find (2x+3)(x-5).

(3) If we know the roots or zeros, we can rewrite the quadratic in factored form. The example `x^2-8x+13` is irreducible in the rationals. (The discriminant ` Delta =(-8)^2-4(1)(13)=12` is not a perfect square.) We can use completing the square or the quadratic formula to find the roots:

(a) a=1, b=-8, c=13, so using the quadratic formula`"if "ax^2+bx+c=0" then "x=(-b pm sqrt(b^2-4ac))/(2a)`, we get

`x=4+sqrt(3),4-sqrt(3)` so we can factor as `(x-4-sqrt(3))(x-4+sqrt(3))`

(b) If `x^2-8x+13=0 ==> x^2-8x=-13 ==> x^2-8x+16=-13+16`, then `(x-4)^2=3 ==> x-4=pm sqrt(3) ==> x=4pm sqrt(3)`, as before.