Factoring trinomials with a leading coefficient of 1

Question

5W^4-35W^3+50W^2

Answer 1

The coefficients of 5w⁴ - 35w³ + 50w² are 5, -35, and 50.

These three numbers are multiples of 5.

So we can begin by factoring out the 5:

5w⁴ - 35w³ + 50w2 = 5(w⁴ - 7w³ + 10w²)

Now we've got a trinomial w⁴ - 7w³ + 10w² with smaller coefficients than those of the original trinomial 5w⁴ - 35w³ + 50w², and the leading has been reduced all the way down to 1. (We could write it as 1w⁴ - 7w³ + 10w².)

The exponents of w⁴ - 7w³ + 10w² are 4, 3, and 2.

These three numbers are greater than or equal to 2.

So we can continue by factoring out the x² as well as the 5:

5w⁴ - 35w³ + 50w² = 5w²(w² - 7w + 10)

Now we've got a trinomial w² - 7w + 10 that's simple enough to factor.

One pair of factors of 10 are 2 and 5. Unfortunately, 2 + 5 = 7.

Another pair of factors of 10 are -2 and -5. Fortunately, -2 + (-5) = -7, which is the sum we want.

So we can factor 5w⁴ - 35w³ + 50w² all the way down to the three factors

5w⁴ - 35w³ + 50w² = (5w²)(w - 2)(w - 5)

Answer 2

First see if there is any way you can factor this. Look for similar numbers and see if there is a W in each term as well.

does this help?

5W²(W²-7W+10) Find the factors of 10.

5W²(W-5)(W-2)