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Binomial Times Trinomial
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and the solution set X is {0, 6, 8} A new root shows up this time: x = 0! The fact that one of the roots is 0 caused us to inadvertently divide by 0 when we divided the equation through by x This blinded us to the existence of that root
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Binomial Times Trinomial
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When a cubic can be expressed as a binomial multiplied by a trinomial, the equation is in binomial-trinomial form (Actually, I ve never seen that expression used in other texts But it s easy to remember, don t you think ) A cubic in this form is not particularly difficult to solve for real roots The technique shown in this section will also reveal the complex-number roots of a cubic equation, if any such roots exist
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Binomial-trinomial form Suppose that a1 and a2 are nonzero real numbers Also suppose that b1, b2, and c are real numbers, any or all of which can equal 0 The binomial-trinomial form of a cubic equation in the variable x can be written as follows:
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(a1x + b1)(a2x2 + b2x + c) = 0 Here are some examples of cubics in the binomial-trinomial form: ( 4x 3)( 7x 2 + 6x 13) = 0 (3x + 5)(16x 2 56x + 49) = 0 (3x)(4x 2 7x 10) = 0 ( 21x + 2)(3x 2 14) = 0 In the third case above, the stand-alone constant is 0 in the binomial In the fourth case, the coefficient of x is 0 in the trinomial
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Multiplying out Let s take a specific example of a cubic in binomial-trinomial form and multiply it out Here s a good one, with plenty of sign changes to make it interesting:
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( 4x 3)( 7x 2 + 6x 13) = 0 Using the expanded product of sums rule, we obtain 28x 3 24x 2 + 52x + 21x 2 18x + 39 = 0 Consolidating the terms for x 2 and x, we get 28x 3 3x 2 + 34x + 39 = 0
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What are the real roots The process of finding the real roots of a cubic in the binomial-trinomial form is straightforward, as long as all the coefficients and constants are real numbers First, we manufacture a first-degree equation from the binomial, setting it equal to 0 In the general form above, that would be
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a1x + b1 = 0 which solves to x = b1/a1 This will always give us one real root for the cubic After that, we set the trinomial equal to 0, obtaining a quadratic equation In the general form shown above, we get a2x 2 + b2x + c = 0 We can find the real roots of this equation, if any exist, using techniques we ve already learned for solving quadratics (I like to use the quadratic formula, because it always works! Also, if the root or roots are complex but not real, the quadratic formula will produce them) Expressed for the above general equation, the quadratic formula is x = [ b2 (b22 4a2c)1/2] / (2a2)
Are you confused
You ve learned that a quadratic equation can have two different real roots, or only one real root, or none at all How about cubics You ve already seen an example of a cubic with three real roots You re about to see that a cubic equation in the binomial-trinomial form with real coefficients and a real constant can have two real roots Then you ll discover that a cubic in the binomial-trinomial form with real coefficients and a real constant can have only one real root, along with two others that are complex Okay, you say Then you ask, How about no real roots The answer: Any cubic in the binomialtrinomial form with real coefficients and a real constant always has at least one real root That s because a first-degree equation can always be created from the binomial factor, and that equation always has a real solution We can take this statement further: A cubic equation, no matter what the form, has at least one real root if all the coefficients and constants are real numbers
Here s a challenge!
Find the real roots of the following cubic equation using the method described in this section, and state the real solution set X (3x + 5)(16x 2 56x + 49) = 0
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