Saved Bookmarks
| 1. |
Write the second degree polynomials. given below as the product of two first degree polynomials. Find also the solutions of the equation p(x) = 0 in each.i. p(x) = x2 – 7x+12 ii. p(x) = x2 + 7x + 12 iii. p (x) = x2 – 8x +12 iv. p(x) = x2 + 13x +12 v. p (x) = x2 + 12x – 13 vi. p (x) = x2 – 12x – 13 |
|
Answer» i. p (x) = x2 – 7x + 12 a + b = –7, ab = 12 a = –3, b = –4 x2 – 7x + 12 = (x – 3) (x – 4) x2 – 7x + 12 = 0 (x – 3) (x – 4) = 0 x – 3 = 0, x – 4 = 0 x = 3, x = 4 ii. p(x) = x2 + 7x + 12 a + b = 7, ab= 12 a = 3, b = 4 x2 + 7x + 12 = (x + 3) (x + 4) x2 + 7x + 12 = 0 (x + 3)(x + 4) = 0 x = 3, x = 4 iii. p(x) = x2 - 8x + 12 a + b = 8, ab= 12 a = 6, b = –2 x2 – 8x + 12 = (x – 6) (x – 2) (x – 6) (x – 2) = 0 x = 6, x = 2 iv. p(x) =x2 + 13x + 12 a + b = 13, ab = 12 a =12, b=1 (x +13x + 12) = (x + 12) (x + 1) x2 +13x + 12 = 0 (x+ 12) (x+ 1) = 0 x+ 12 = 0, x+ 1 = 0 x = -12 ,x = –1 v. p(x) = x2 + 12x – 13 a = –13, b = 1 x2 + 12x – 13 = (x + 13)(x – 1) x + 13 = 0, x – 1 =0 x = –13, x= 1 . vi. p(x) = x2 – 12x – 13 x2 – 12x – 13 = (x – a) (x – b) = x2 – (a + b) x + ab a + b= 12 ab = –13 (a – b)2 =(a + b)2 – 4ab = (12)2 – 4x – 13 = 196 a – b = 14 a + b= 12 a= 13; b = –1 x2 – 12x – 13 = (x – 13)(x + 1) x – 13 = 0, x + 1 =0 x= 13 ,x = –1 |
|