Solve the following equation: x^(3) - 6x^(2) + 11x - 6 = 0
x3 - 6x2 + 11x - 6 = 0
Let (x-a)(x-b)(x-c) = x3 - 6x2 + 11x - 6=> abc = -6 and a + b + c = -6
From abc=-6, find the possibilities of their values: (1,2,-3), (1,-2,3), (-1,2,3), (-1,-2,-3), (1,1,-6), (1,-1,6), (-1,1,6) & (-1,-1,-6)
Then find the sum of each set of brackets (making sure you do this for all of them):e.g. 1 + 2 + (-3) = 0 and (-1) + (-2) + (-3) = -6
Now, using a + b + c = -6, eliminate the values which do not work.
In this case, the only set that work are (-1,-2,-3) .
So, giving us (x-1)(x-2)(x-3) = 0
To find x solve each bracket:=> x-1=0, x-2=0 & x-3=0
=> x=1, x=2 & x=3
Hence, these are the 3 values of x.
[=> means "implies"]
EH
Related Maths A Level answers
All answers ▸Use the substitution u=2+ln(t) to find the exact value of the antiderivative of 1/(t(2+ln(t))^2)dt between e and 1.
A level Maths question - The graph of y=2sin(2x)+1 is rotated 360 degrees about the x-axis to form a solid. Find the volume enclosed by the curve, the co-ordinate axes and the line x=pi/2
Two points have coordinates (1,-6) and (-2,3). Find the equation of the line which joins them, and their midpoint.
Use the double angle formulae and the identity cos(A+B)≡cos(A)cos(B)−sin(A)sin(B) to obtain an expression for cos 3x in terms of cos x only