let t1, t2, t3, t4 be the terms in the binomials, the product will look like this:

(t1 + t2)(t3 + t4)

Step 2: Find the factors t1 and t3 that go in the first positions. The product of these factors will need to equal b^2.

(t1)(t3) = b^2

To get the b^2 term, let t1= b and t3=b. So the binomial product will look like this now:

(b + t2)(b + t4)

Step 3: Find the factors t2 and t4 that go in the last positions. The product of these factors will need to equal -160.

(t2)(t4) = -160

and at the same time

(t1)(t4) + (t2)(t3) = -12b

Substituting t1= b and t3= b in the previous equation:

b(t4) + (t2)b = -12b

Try t2= 8 and t4= -20, since (t2)(t4) = -160

Try t2= 8 and t4= -20 in the preceding equation

b(-20) + (8)b = -12b

which is what we needed.

So, the polynomial factors are:

(t1 + t2)(t3 + t4)

where t1=b, t3= b, t2=8, t4=-20

(b + 8)(b – 20)

Note: You must have a value on the right side of the equal sign to find actual solutions. Since there is nothing on the right side, you’re done.

But I can’t see how to solve it.

If it was b^2-12b-160=0

Then b=20 or -8

However, the polynomial can be simplified by factoring:

(b-20)(b+8)

Best of luck!

Therefore eaither b=-8 or b=20