1. Partial fractions
Repeated factors

2. 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

3. 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=− then =(0)+𝐵)
gives 𝐵=3
Let 𝑥=2 then =𝐴(1)+𝐵
gives 𝐴=4
4𝑥−1 𝑥−1 2 = 4 (𝑥−1) 𝑥−1 2

4. 11 𝑥 2 +14𝑥+5=𝐴(𝑥+1)(2𝑥+1)+ 𝐵(2𝑥+1)+ 𝐶 𝑥+1 2
WB7b write b) 𝑥 2 +14𝑥+5 𝑥+1 2 (2𝑥+1) in partial fractions
= 𝐴 (𝑥+1) + 𝐵 𝑥 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 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=− then −14+5= 0 +𝐵 −1 +(0)
gives 𝐵=-2
Let 𝑥=− then − = 𝐶 1 4
gives 𝐶=3
Let 𝑥= then =𝐴 1 +𝐵 1 +𝐶 1
gives 𝐴=4
11 𝑥 2 +14𝑥+5 𝑥+1 2 (2𝑥+1) = 4 (𝑥+1) − 2 𝑥 𝑥+1

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

6. Partial fractions
Top-Heavy fractions

7. 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.

8. 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 𝑥 2 −3𝑥−2
6𝑥−8
3 𝑥 2 −3𝑥−2 (𝑥−1)(𝑥−2) = 𝑥−8 (𝑥−1)(𝑥−2)

9. Now equate the numerators
WB8 (cont) 𝑥 2 −3𝑥−2 (𝑥−1)(𝑥−2) = 𝑥−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= gives B = 4
Let 𝑥=1 gives 𝐴=2
3 𝑥 2 −3𝑥−2 (𝑥−1)(𝑥−2) = 𝑥−1 + 4 𝑥−2

10. 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= then −3−2= 0 +𝑄 −1 +(0)
gives 𝑄=2
Let 𝑥= then −6−2= 𝑅(1)
gives 𝑅=4
Let 𝑥= 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

11. 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=− then −40−10= 0 +𝑄 −7 +(0)
gives 𝑄=2
Let 𝑥= then −10= 𝑅( 7 3 )
gives 𝑅=−1
Let 𝑥= 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

12. 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 –

13. END