The partial factorization of x2 – x – 12 is modeled with algebra tiles.

An algebra tile configuration. 4 tiles are in the Factor 1 spot: 1 is labeled + x, 3 are labeled +. 1 tile is in the Factor 2 spot and is labeled + x. 20 tiles are in the Product spot: 1 is labeled + x squared, 3 are labeled + x, the 4 tiles below + x squared are labeled negative x, and the 12 tiles below the + x tiles are labeled negative.
Which unit tiles are needed to complete the factorization?

Respuesta :

Answer:

x² – x – 12 = (x – 4)(x + 3)

Step-by-step explanation:

Identify two numbers that add to -1 and multiply to -12, let's call them p and q.

So ax² + bx + c = (x + p)(x + q)

pq = c

p + q = b.

It is easier to find these numbers by finding factors of -12.

This can be done by splitting the number up until all the numbers are prime.

-12 → 6 × -2 or -6 × 2 → -(3 × 2 × 2)

There can only be two numbers so the only options we have are 6 and -2, -6 and 2, 3, and -4, or -3 and 4.

We can eliminate them by adding them up.

6 + -2 = 4 ≠ -1 so that can't be it.

-6 + 2 = -4 ≠ -1 so that can't be it either.

-3 + 4 = 1 ≠ -1

therefore p and q are 3 and -4 because 3 + -4 = -1.

so x² – x – 12 = (x – 4)(x + 3)

p = -4, and q = 3.

(x – 4)(x + 3) = x(x + 3) – 4(x + 3) = x² + 3x – 4x + 12 = x² – x – 12