(x+1)(x+5)(x+2)
x3 + x2 - 17x + 15 = (x - 1)(x - 3)(x + 5). Thus, the zeros are 1, 3, and -5. All three zeros are rational.
18x2 + 15x + 2 = (3x + 2)(6x + 1)
3x2 + 17x + c = 0, rearranging gives c = -3x2 - 17x
108x2-102x+24 Simplified by dividing all terms by 6: 18x2-17x+4 When factorised: (9x-4)(2x-1) With a question like this using the quadratic equation formula will help.
(x+1)(x+5)(x+2)
x3 + x2 - 17x + 15 = (x - 1)(x - 3)(x + 5). Thus, the zeros are 1, 3, and -5. All three zeros are rational.
(x + 1)(x + 3)(x - 5)
18x2 + 15x + 2 = (3x + 2)(6x + 1)
3x2 + 17x + c = 0, rearranging gives c = -3x2 - 17x
Dividend: 4x^4 -x^2 +17x^2 +11x +4 Divisor: 4x +3 Quotient: x^3 -x^2 +5x -1 Remainder: 7
108x2-102x+24 Simplified by dividing all terms by 6: 18x2-17x+4 When factorised: (9x-4)(2x-1) With a question like this using the quadratic equation formula will help.
It is 1.17x
17x-3 = 14
17x + 9
17x + 13
6x2-17x+7 = (2x-1)(3x-7) when factored