Saved Bookmarks
| 1. |
Using factor theorem, factorize each of the following polynomial:x3 - 23x2 + 142x - 120 |
|
Answer» Let, f (x) = x3 - 23x2 + 142x - 120 The factors of the constant term – 120 are \(\pm\)1, \(\pm\)2, \(\pm\)3, \(\pm\)4, \(\pm\)5, \(\pm\)6, \(\pm\)7, \(\pm\)8, \(\pm\)9, \(\pm\)10, \(\pm\)11,\(\pm\)12, \(\pm\) 15, \(\pm\) 20, \(\pm\) 24, \(\pm\) 30, \(\pm\) 40, \(\pm\) 60, and \(\pm\) 120 Putting x = 1, we have f (1) = (1)3 – 23 (1)2 + 142 (1) – 120 = 1 – 23 + 142 – 120 = 0 So, (x – 1) is a factor of f (x) Let us now divide f (x) = x3 - 23x2 + 142x - 120 by (x - 1) to get the other factors of f (x) Using long division method, we get x3 - 23x2 + 142x - 120 = (x – 1) (x2 – 22x + 120) x2 – 22x + 120 = x2 – 10x – 12x + 120 = x (x – 10) – 12 (x – 10) Hence, x3 - 23x2 + 142x - 120 = (x – 1) (x - 10) (x - 12) |
|