Question

4a^2-17a+4 factor polynomial

Answer 1

(4x - 1)(x - 4)

Explanation:

Using what we did from the previous question, we get (ax + b)(cx + d), which is (ac)x^2 + (ad + bc)x + bd. Setting up equations, we have ac = 4

ad + bc = -17

bd = 4.

Let's guess and check a few values. First, because -17 is odd, and 4 can be 2*2, we know it cannot be 2*2 because that would results in an even number in the coefficient of x. Now, we know a is 4 and c is 1 (or vice versa). Using he same logic, b and d cannot be 2, so they are 4 and 1 or -4 and -1. Because ad + bc = -17, 4d + c = -17, so d = -4 and c = -1, and the final answer is (4x - 1)(x - 4).
`\blacksquare`

Answer 2

(4a - 1 ) (a- 4 )

Explanation:

4a^2-17a+4

This one is a bit tricker

We could split the 4a term, but we notice that we need to get to 17 in the middle so my guess is that I will need to leave it as 4

(4a ) (a )

We notice that the 17 is negative and the 4 at the end is positive so both terms will need to be negative

(4a - ) (a- )

The only way to get to add to -17 and multiply to +4

4*-4 + 1*-1 = -16-1 =-17

And -4*-1 = 4

We put the -4 so it will multiply by the 4a

(4a - 1 ) (a- 4 )