InterviewSolution
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InterviewSolution
| 1. |
Factorise: 2xcube-3x-17x+30 |
| Answer» Let f(x) = 2x3 - 3x2 - 17x + 30 be the given polynomial. The factors of the constant term +30 are {tex}\\pm 1, \\pm 2, \\pm 3, \\pm 5, \\pm 6, \\pm 10, \\pm 15, \\pm 30{/tex}. The factor of coefficient of x3 is 2. Hence, possible rational roots of f(x) are:{tex}\\pm 1, \\pm 3, \\pm 5, \\pm 15, \\pm \\frac{1}{2}, \\pm \\frac{3}{2}, \\pm \\frac{5}{2}, \\pm \\frac{{15}}{2}{/tex}We have f(2) = 2(2)3 - 3(2)2 - 17(2) + 30= 2(8) - 3(4) - 17(2) + 30= 16 - 12 - 34 + 30 = 0And f(-3) = 2(-3)3 - 3(-3)2 - 17(-3) + 30= 2(-27) - 3(9) - 17(-3) + 30= -54 - 27 + 51 + 30 = 0So, (x - 2) and (x + 3) are factors of f(x).{tex}\\Rightarrow{/tex} x2 + x - 6 is a factor of f(x).Let us noe divide f(x) = 2x3 - 3x2 - 17x + 30 by x2 + x - 6 to get the other factors of f(x).Factors of f(x).By long division, we have{tex}\\therefore{/tex} 2x3 - 3x2 - 17x + 30 = (x2 + x - 6)(2x - 5){tex}\\Rightarrow{/tex} 2x3 - 3x2 - 17x + 30 = (x - 2)(x + 3)(2x - 5)Hence, 2x3 - 3x2 - 17x + 30 = (x - 2)(x + 3)(2x - 5) | |