WEBVTT
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Express π₯ minus two over π₯ times π₯ minus three in partial fractions.
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We have the algebraic fraction π₯ minus two over π₯ times π₯ minus three, and we want to write it in partial fractions.
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That means decomposing it as the sum of two fractions, where the denominators of our two fractions are the factors of the original denominator.
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Weβre lucky that our denominator is already factored for us.
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We have π₯ times π₯ minus three, and so the denominators of our two fractions in our decomposition are going to be π₯ and π₯ minus three.
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The numerators of these two fractions are constants that we need to find.
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Weβll call them π΄ and π΅.
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This is the form of our partial fraction decomposition.
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We just need to find the values of π΄ and π΅ now and weβll be done, and we find the values of π΄ and π΅ by doing some algebra.
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We multiply both sides by π₯ times π₯ minus three, and we notice that we can cancel the denominator of each faction on the right-hand side with a factor of its numerator, so the π₯s cancel in the first fraction and the π₯ minus threes cancel in the second fraction.
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So we are left with π΄ times π₯ minus three plus π΅π₯ on the right-hand side.
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We can expand out the brackets on the right-hand side β π΄ times π₯ minus three becomes π΄π₯ minus three π΄ β and then combine like terms, so π΄π₯ plus π΅π₯ becomes π΄ plus π΅ times π₯.
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And as the equation weβre left with β π₯ minus two equals π΄ plus π΅ times π₯ minus three π΄ β must hold for all values of π₯, we can compare the coefficients.
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The coefficient of π₯ on the left-hand side is one and on the right-hand side is π΄ plus π΅, and these coefficients must be equal, so π΄ plus π΅ must be one.
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And for the same reason, the constant term on the left-hand side, negative two, must be equal to the constant term on the right-hand side, negative three π΄.
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From this second equation, we can see that π΄ is two over three, and we can use this value in our first equation β π΄ plus π΅ equals one β to get that two-thirds plus π΅ equals one and hence that π΅ equals one-third.
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Now that we have the values of π΄ and π΅, we can just put them into the expression we have.
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π₯ minus two over π₯ times π₯ minus three is two-thirds over π₯ plus one-third over π₯ minus three.
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And of course, we would prefer not to have a fraction in the numerator, so two-thirds over π₯ becomes two over three π₯ and a third over π₯ minus three becomes one over three times π₯ minus three.
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And we get our final answer: π₯ minus two over π₯ times π₯ minus three expressed in partial fractions is two over three π₯ plus one over three times π₯ minus three.
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And you might like to check that this is true for all π₯s just by substituting different values of π₯ into this equation, not including π₯ equals zero or π₯ equals three when both sides of the equation are undefined.