Factorise 2x^3-5x^2-19x+42
(using trial and error long method)
Answers
We can use synthetic division to look for factors:
The possible zeroes are the (factors of 42)/(factors of 2)
Possible zeroes: ±{1, 1/2, 3, 3/2, 6, 7, 7/2, 14, 21, 10 1/2, 42}
If (x+1) is a factor then x = -1 is a zero:
-1 | 2 -5 -19 42
| -2 7 12
----------------------
2 -7 -12 54 This has remainder ≠0 so not a factor
I also did (x-1), (x-2), (x+2) until I got to (x+3)
-3 | 2 -5 -19 42
| -6 33 -42
--------------------------
2 -11 14 0 This has remainder of zero so -3 is a zero
meaning that x+3 is a factor
2x^3-5x^2-19x+42 = (x+3)(2x2 - 11x + 14)
We can factor the 2x2 - 11x + 14 by using grouping:
Looking at the coefficients of the x2 term and the constant
we have 2 & 14... 2(14) = 28. Look for factors of 28 that
add to the coefficient of the x term ... -11
(-4)(-7) = 28 and -4-7 = -11.
Replace -11x with -4x - 7x and factor by grouping
2x2 - 11x + 14 = 2x2 - 4x - 7x + 14
= 2x(x - 2) -7(x - 2) = (2x-7)(x-2)
Our final answer:
2x3-5x2-19x+42 = (x+3)(2x-7)(x-2)