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By: Lawrence, Benjamin, Lola, and Liz Trig 1 PD 8 . Partial Fractional Decomposition . What is Partial Fractional Decomposition? . It’s a process that takes one fraction and expresses it as the sum or difference of two other fractions. Step one .
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By: Lawrence, Benjamin, Lola, and Liz Trig 1 PD 8 Partial Fractional Decomposition
What is Partial Fractional Decomposition? • It’s a process that takes one fraction and expresses it as the sum or difference of two other fractions
Step one • Factor the denominator and rewrite it as A over one factor and B over the other. • You do this because you want to break the fraction into two. The process unfolds as follows:
Step Two • Multiply every term you’ve created by the factored denominator and then cancel. • This equals 11x + 21 = A(x + 6) + B(2x – 3).
Step Three • Distribute A and B. • 11x + 21 = A(x + 6) + B(2x – 3). • This gives you 11x + 21 = Ax + 6A + 2Bx – 3B.
Step Four • On the right side of the equation only, put all terms with an x together and all terms without it together. • Rearranging gives you 11x + 21 = Ax + 2Bx + 6A – 3B.
Step Five • Factor out the x from the terms on the right side. • You now have 11x + 21 = x (A + 2B) + 6A – 3B.
Step Six • Create a system out of this equation by pairing up terms. • For an equation to work, everything must be in balance. • Because of this fact, the coefficients of x must be equal and the constants must be equal. • If the coefficient of x is 11 on the left and A + 2B on the right, you can say that 11 = A + 2B is one equation. • Constants are the terms with no variable, and in this case, the constant on the left is 21. On the right side, 6A – 3B is the constant (because there is no variable attached) and so 21 = 6A – 3B.
Step Seven • Solve the system, using either substitution or elimination. • In this example, you use elimination in this system. -6A-12B=-66 6A-3B=21 -6 -15B=-45 B=3 and A=5
Step Eight= Solution • Write the solution as the sum of two fractions.