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Although there are formal methods, the simplest is to see if the cubic can be factorised. If so, you'll have a linear factor and a quadratic, both of which are easy to solve. Alternatively, you could try solving it graphically. Another possibility is to solve it numerically. The derivative of a cubic, of the form f(x) = ax3 + bx2 + cx + d is f'(x) = 3ax2 + 2bx + c. Make an initial estimate x1. An improved estimate is x2 = x1 - f(x1)/f'(x1). Use x2 to make next estimate, and so on.
(x2 - 9) / (x2 + 3) does not have a simple solution. Perhaps the question you were actually looking to have answered was:(x2 - 9) / (x + 3)Which is equal to (x - 3). This is called a difference of squares, where any time you see a term in the format:(x2 - b2)You know that it is equal to (x + b)(x - b). Of course, (x + b)(x - b) / (x + b) will be equal to (x - b), giving you the answer above.If on the other hand, you are indeed looking for the answer to (x2 - 9) / (x2 + 3), you could try long division. That will give you 1 with a remainder of negative twelve, or:1 + 12/(x2 + 3)Alternately, if you have a value for the term, you can solve the equation:(x2 - 9) / (x + 3) = ax2 - 9 = ax + 3ax2 - ax = 9 + 3ax2 - ax + (a/2)2 = 9 + 3a + (a/2)2 ← this is called "completing the square".(x - (a/2))2 = 9 + 3a + (a/2)2x - (a/2) = [9 + 3a + (a/2)2]1/2x = [9 + 3a + (a/2)2]1/2 + a/2You can then plug in whatever value is given for 'a'. For example, if it's equal to zero:x = [9 + 3(0) + (0/2)2]1/2 + 0/2x = (9 + 0 + 0)1/2 + 0x = 91/2x = 3Note that x can not be equal to negative three, as that would cause division by zero in the original equation.
ax2 + bx + c is the general formula for a quadratic trinomial.If a = 1 it is easier. Find 2 integers whose product = c AND whose sum = b.The 2 number must do both. call thos numbers m and nThe factors become (a+m)(a+n)ex: x2 - x - 12c = -12, has many factors -1 x 12, -2 x 6, -3 x 4, -4x3, -6x2, and -1x12compare these to b = -1. the only ones that have a sum of -1 are -4 and 3so the factors are (x-4)(x+3) --- answerBut if a is not equal to 1step1. Multiply a and c find a new number, lets call it d. Use d instead of c like above to find factors and look for a sum equal to b. just like above but you are factoring d NOT c. lets call these factors m and n just like before.step 2. Replace bx with mx + nx Now you have 4 terms instead of 3ax2 + bx + c becomes ax2 + mx + nx + c group in twos using parenthesis( ax2 + mx) + (nx + c)step 3. Look at each group separately and get a GCF.step 4, Look at the parenthesis after each above GCF. If they are not the same the original trinomial was prime (or you made a mistake). If they are the same then put the GCF's into a parenthesis multiplied by the parenthesis that was the same. This is the answer. This sounds complicated, but it is easier to do than to explain using a computer.Ex: Factor 2x2 +11x +12Step 1 --- 2x12 = 24, Factors are 1x24, 2x12, 3x8, 4x6 and some negatives which we don't need THIS TIME. Look for the sum that = b = 11 ... 3+8 = 11Step 2 --- replace b ---- 2x2 +11x +12 = 2x2 + 8x + 3x +12group ---- ( 2x2 + 8x) + ( 3x +12)Step 3 --- GCF --- 2x(x + 4) + 3(x + 4) ----- notice parenthesis is same.Step 4 ---- Both GCF's make a new group and same parenthesis is used once(2x + 3)(x+4)This is the answer.