Introduction Learn to Solve Cubic Equations In mathematical terms, all cubic equations have either one root or three real roots. The general cubic equation is, ax3+ bx2+ cx+d= 0 The coefficients of a, b, c and d are real or complex numbers with a not equals to zero (a ≠0). It must have the term x3 in it, or else it will not be a cubic equation.
Introduction The coefficients of a, b, c and d are real or complex numbers with a not equals to zero (a ≠0). It must have the term x3 in it, or else it will not be a cubic equation. But any or all of b, c and d can be zero. The examples of cubic equations are, No 1, x3+ 3a3+ 3 a2 + a3– b=0 No 2, 4x3+ 57=0 No 3, x3+ 9x=0
Strategy to Solve Cubic Equation Unlike quadratic equation which may have no real solution; a cubic equation always has at least one real root. The prior strategy of solving a cubic equation is to reduce it to a quadratic equation, and then solve the quadratic by usual means, either by factorizing or using a formula. Always try to find the solution of cubic equations with the help of the general equation, ax3+ bx2+ cx+d= 0
Strategy to Solve Cubic Equation A cubic equation should, therefore, must be rearranged into its standard form, For example, x2+ 4x-1 = 6/x
Strategy to Solve Cubic Equation Step 1 You can see the equation is not written in standard form, you need to multiply the ‘x’ to eliminate the fraction and get cubic equation, after doing so, you will end up with x3+ 4x2– x = 6
Strategy to Solve Cubic Equation Step 2 Then you subtract 6 from both sides in order to get ‘0’ on the right side, so you will come up, x3+ 4x2– x- 6 = 0
Solving Cubic Equations with the help of Factor Theorem What is factor theorem? If you divide a polynomial p(x) by a factor x – a of that polynomial, then, you will end up with zero as the remainder, p(x) = (x – a)q(x) + r(x)
Solving Cubic Equations with the help of Factor Theorem If x – a is indeed a factor of p(x), then the remainder after division by x – a will be zero. p(x) = (x – a)q (x) Here is a problem, x3– 5x2– 2x+24 = 0 With x= – 2 a solution.
Solving Cubic Equations with the help of Factor Theorem If x – a is indeed a factor of p(x), then the remainder after division by x – a will be zero. p(x) = (x – a)q (x) Here is a problem, x3– 5x2– 2x+24 = 0 With x= – 2 a solution.
Solving Cubic Equations with the help of Factor Theorem Factor theorem says that if x = – 2 is a solution of this equation, then x+2 is a factor of this whole expression.
Solving Cubic Equations with the help of Factor Theorem Step 1 First, you need to look at the coefficients of the original cubic equation, which are 1, -5, -2 and 24.
Solving Cubic Equations with the help of Factor Theorem Step 2 Now multiply number (1) that just brought down, by the known root -2, as a result is -2, you mention the result in the other line, like
Solving Cubic Equations with the help of Factor Theorem Step 3 The numbers in the second column are added, so giving us,
Solving Cubic Equations with the help of Factor Theorem Step 4 Then recently written number 7 is multiplied by the known root, – 2, As 14 comes as a result, you need to write it down on the second row over the line,
Solving Cubic Equations with the help of Factor Theorem Step 5 Like previously the numbers in this column added, (14 – 2 = 12)
Solving Cubic Equations with the help of Factor Theorem Step 6 And you need to go on with the process,
Solving Cubic Equations with the help of Factor Theorem Step 7 When you have zero at the bottom row, it gives the confirmation that x = – 2 is a root of the original cubic. At this stage, you got the first three numbers in the bottom row as the coefficients in the quadratic, x2– 7x+12 Hence, you reduced your cubic to,
Solving Cubic Equations with the help of Factor Theorem (x+2)(x2 – 7x + 12) =0 Step 8 After applying the quadratic term, the equation comes like this, (x +2) (x – 3) (x – 4) = 0 Resulting, you get the solution as x = -2 or 3 or 4.
Solving Cubic Equations with the help of Factor Theorem Another Example: The equation is, x3– 7x-6=0 Step 1 You can simply try x = – 1, after putting the value of x, you will get, (-1)3 – 7(-1) -6
Solving Cubic Equations with the help of Factor Theorem Step 2 After applying the synthetic division, like above example, you will take the coefficients of the original cubic equation, which are 1, 0, -7 and -6, you need to write down the know root x = -1 to the right of the vertical line, giving us,
Solving Cubic Equations with the help of Factor Theorem Step 3 Multiply the brought down number 1 by the known root x = -1, and put down the result (1) at the second row, like this,
Solving Cubic Equations with the help of Factor Theorem Step 4 The numbers of the second column are added to the first column, giving us,
Solving Cubic Equations with the help of Factor Theorem Step 5 As you add more numbers to the second column by following the synthetic division process, you will come with,
Solving Cubic Equations with the help of Factor Theorem Step 6 Hence, the cubic reduced to quadratic, (x+1)(x2-x- 6) =0 The factorized result is, (x +1)(x – 3)(x + 2) = 0 You can get three solutions to the cubic equation are x = -2, -1 or 3
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