Partial fractions Repeated factors

Algebra: Partial fractions KUS objectives BAT put expressions into partial fractions where the denominator has repeated factors or the original fraction is improper (top heavy) Starter: xx

4𝑥−1=𝐴(𝑥−1)+𝐵 WB7a write a) 4𝑥−1 𝑥−1 2 in partial fractions = 𝐴 (𝑥−1) + 𝐵 𝑥−1 2 Make the denominators the same 4𝑥−1 𝑥−1 2 = 𝐴(𝑥−1) (𝑥−1)(𝑥−1) + 𝐵 𝑥−1 2 Now equate the numerators 4𝑥−1=𝐴(𝑥−1)+𝐵 Let 𝑥=−1 then 3=(0)+𝐵) gives 𝐵=3 Let 𝑥=2 then 7=𝐴(1)+𝐵 gives 𝐴=4 4𝑥−1 𝑥−1 2 = 4 (𝑥−1) + 3 𝑥−1 2

11 𝑥 2 +14𝑥+5=𝐴(𝑥+1)(2𝑥+1)+ 𝐵(2𝑥+1)+ 𝐶 𝑥+1 2 WB7b write b) 11 𝑥 2 +14𝑥+5 𝑥+1 2 (2𝑥+1) in partial fractions = 𝐴 (𝑥+1) + 𝐵 𝑥+1 2 + 𝐶 2𝑥+1 Make the denominators the same 11 𝑥 2 +14𝑥+5 𝑥+1 2 (2𝑥+1) = 𝐴(𝑥+1)(2𝑥+1) 𝑥+1 2 (2𝑥+1) + 𝐵(2𝑥+1) 𝑥+1 2 (2𝑥+1) + 𝐶 𝑥+1 2 𝑥+1 2 (2𝑥+1) Now equate the numerators 11 𝑥 2 +14𝑥+5=𝐴(𝑥+1)(2𝑥+1)+ 𝐵(2𝑥+1)+ 𝐶 𝑥+1 2 Let 𝑥=−1 then 11−14+5= 0 +𝐵 −1 +(0) gives 𝐵=-2 Let 𝑥=− 1 2 then 11 4 − 14 2 +5= 0 + 0 +𝐶 1 4 gives 𝐶=3 Let 𝑥=0 then 5=𝐴 1 +𝐵 1 +𝐶 1 gives 𝐴=4 11 𝑥 2 +14𝑥+5 𝑥+1 2 (2𝑥+1) = 4 (𝑥+1) − 2 𝑥+1 2 + 3 2𝑥+1

3 𝑥 2 +18=𝐴 2+𝑥 2 +𝐵(1−2𝑥)(2+𝑥)+ 𝐶(1−2𝑥) WB7c write c) 3 𝑥 2 +18 (1−2𝑥) 2+𝑥 2 in partial fractions Include a repeated factor term when usig partial fractions 3 𝑥 2 +18 (1−2𝑥) 2+𝑥 2 = 𝐴 (1−2𝑥) + 𝐵 2+𝑥 + 𝐶 2+𝑥 2 Now equate the numerators 3 𝑥 2 +18=𝐴 2+𝑥 2 +𝐵(1−2𝑥)(2+𝑥)+ 𝐶(1−2𝑥) Let 𝑥=−2 then 12+18= 0 + 0 +𝐶(5) gives C= 6 Let 𝑥= 1 2 then 3 4 +18=𝐴 25 4 + 0 + 0 gives 𝐴=3 Let 𝑥=0 then 18=𝐴 4 +𝐵 2 +𝐶 1 gives 𝐵=0 3 𝑥 2 +18 (1−2𝑥) 2+𝑥 2 = 3 (1−2𝑥) + 6 2+𝑥 2

Partial fractions Top-Heavy fractions

Dealing with Improper Fractions The ‘degree’ of a polynomial is the highest power, e.g. a quadratic has degree 2. An algebraic fraction is improper if the degree of the numerator is at least the degree of the denominator.        

Divide the numerator by the denominator to find the ‘whole’ part WB8 write 3 𝑥 2 −3𝑥−2 (𝑥−1)(𝑥−2) in partial fractions Remember, Algebraically an ‘improper’ fraction is one where the degree (power) of the numerator is equal to or exceeds that of the denominator 3 𝑥 2 −3𝑥−2 (𝑥−1)(𝑥−2) = 3 𝑥 2 −3𝑥−2 𝑥 2 −3𝑥+2 3 𝑥 2 −9𝑥−6 3 Divide the numerator by the denominator to find the ‘whole’ part 𝑥 2 −3𝑥+2 3 𝑥 2 −3𝑥−2 6𝑥−8 3 𝑥 2 −3𝑥−2 (𝑥−1)(𝑥−2) = 3+ 6𝑥−8 (𝑥−1)(𝑥−2)

Now equate the numerators WB8 (cont) 3 𝑥 2 −3𝑥−2 (𝑥−1)(𝑥−2) = 3+ 6𝑥−8 (𝑥−1)(𝑥−2) Remember, Algebraically an ‘improper’ fraction is one where the degree (power) of the numerator is equal to or exceeds that of the denominator 6𝑥−8 (𝑥−1)(𝑥−2) = 𝐴 𝑥−1 + 𝐵 𝑥−2 Now equate the numerators 6𝑥−8=𝐴(𝑥−2)+𝐵(𝑥−1) Let 𝑥=2 gives B = 4 Let 𝑥=1 gives 𝐴=2 3 𝑥 2 −3𝑥−2 (𝑥−1)(𝑥−2) = 3+ 2 𝑥−1 + 4 𝑥−2

WB8 (Alternative) write 3 𝑥 2 −3𝑥−2 (𝑥−1)(𝑥−2) in partial fractions Write as whole part and partial fractoons 3 𝑥 2 −3𝑥−2 (𝑥−1)(𝑥−2) = 𝑃+ 𝑄 𝑥−1 + 𝑅 𝑥−2 Now equate the numerators 3 𝑥 2 −3𝑥−2=𝑃 𝑥−1 𝑥−2 +𝑄 𝑥−2 +𝑅(𝑥−1) Let 𝑥=1 then 3−3−2= 0 +𝑄 −1 +(0) gives 𝑄=2 Let 𝑥=2 then 12−6−2= 0 + 0 +𝑅(1) gives 𝑅=4 Let 𝑥=0 then −2=𝑃 2 +𝑄 −2 +𝑅(−1) gives 𝑃=3 3 𝑥 2 −3𝑥−2 (𝑥−1)(𝑥−2) =3+ 2 𝑥−1 + 4 𝑥−2

WB9 write 9 𝑥 2 +20𝑥−10 (𝑥+2)(3𝑥−1) in partial fractions Write as whole part and partial fractoons 9 𝑥 2 +20𝑥−10 (𝑥+2)(3𝑥−1) = 𝑃+ 𝑄 𝑥+2 + 𝑅 3𝑥−1 Now equate the numerators 9 𝑥 2 +20𝑥−10=𝑃 𝑥+2 3𝑥−1 +𝑄 3𝑥−1 +𝑅(𝑥+2) Let 𝑥=−2 then 36−40−10= 0 +𝑄 −7 +(0) gives 𝑄=2 Let 𝑥= 1 3 then 9 9 + 20 3 −10= 0 + 0 +𝑅( 7 3 ) gives 𝑅=−1 Let 𝑥=0 then −10=𝑃 −2 +𝑄 −1 +𝑅(2) gives 𝑃=3 9 𝑥 2 +20𝑥−10 (𝑥+2)(3𝑥−1) =3+ 2 𝑥+2 − 1 3𝑥−1

One thing to improve is – KUS objectives BAT put expressions into partial fractions where the denominator has repeated factors or the original fraction is improper (top heavy) self-assess One thing learned is – One thing to improve is –

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