Solve the polynomial
40x^2+2x-65
Notice that for x = 1 the value is -23. The number 23 is a prime number and the only factors are thus 23 and 1 up to signs. So, you can simply the search for the factors of the polynomial a great deal by substituting:
x = 1 + t which yields
40 t^2 + 82 t - 23 =
(10 t + 23) (4 t - 1)
You then substitute back x by putting
t = x - 1:
(10 t + 23) (4 t - 1) =
(10 x + 13) (4 x - 5)
To solve the polynomial 40x^2 + 2x - 65, we can use the fact that for x = 1, the value of the polynomial is -23. From this information, we can determine that the factors of the polynomial must be of the form (10x + a) and (4x + b), where a and b are constants.
Let's substitute x = 1 + t, which means we can write the polynomial as:
40(1 + t)^2 + 2(1 + t) - 65
Simplifying this expression gives us:
40(1 + 2t + t^2) + 2 + 2t - 65
Expanding further:
40 + 80t + 40t^2 + 2 + 2t - 65
Combining like terms:
40t^2 + 82t - 23
We can now factor this polynomial. By inspecting the factored form given, we have:
(10t + 23)(4t - 1)
Substituting back t = x - 1, we get:
(10(x - 1) + 23)(4(x - 1) - 1)
Simplifying this expression gives us:
(10x + 13)(4x - 5)
Therefore, the factored form of the polynomial 40x^2 + 2x - 65 is (10x + 13)(4x - 5).