Question 5:
5 | 12x3 B Cx+D
= A+ +
(2x+1)(2x2 +1) 2x+1 2x2 +1
12x3  A(2x+1)(2x2 +1)+B(2x2 +1)+(Cx+D)(2x+1)
For example: Equating coefficients of x3:A=3
Let x = 0, gives A + B + D = 0
1 2x+2
3− −
2x+1 2x2 +1 | B1
M1*
M1dep*
A1
A1
[5] | 1.1
1.1
1.1
1.1
1.1 | Correct form (possibly implied by correct identity).
Identity without fractions. Follow through their partial fraction expression
with one denominator of 2 x + 1 and the other with 2 x 2 + 1 - both numerators
must contain at least a constant unknown. Some examples for M1 below:
B C x + D
+ s o 1 2 x 3  B ( 2 x 2 + 1 ) + ( C x + D ) ( 2 x + 1 )
2 x + 1 2 x 2 + 1
B C
A + + s o 1 2 x 3  A ( 2 x + 1 ) ( 2 x 2 + 1 ) + B ( 2 x 2 + 1 ) + C ( 2 x + 1 )
2 x + 1 2 x 2 + 1
B C
+ so 12x3 B(2x2 +1)+C(2x+1).
2x+1 2x2 +1
A B x 2 + C x + D
+ s o 1 2 x 3  A ( 2 x 2 + 1 ) + ( B x 2 + C x + D ) ( 2 x + 1 ) .
2 x + 1 2 x 2 + 1
Equates coefficients or substitutes to find an equation involving only their
unknowns. Do not award this mark if only two unknowns in their partial
fractions (so must be at least three unknowns (or implied unknowns)).
Any two (A = 3, B = –1, C = –2, D = –2) unknowns correct from a correct
partial fraction form.
All four unknowns correct – condone stating the correct form of the partial
fractions anywhere together with all four unknowns correctly stated without
necessarily bringing both parts together as a single expression at the end.
Alternative method
Constant term of (A =) 3 | B1 | By polynomial division or inspection.
− 6 x 2 − 6 x − 3 B C x + D
 + and so
( 2 x + 1 ) ( 2 x 2 + 1 ) 2 x + 1 2 x 2 + 1
−6x2 −6x−3B(2x2 +1)+(Cx+D)(2x+1) | − 6 x 2 − 6 x − 3 B C x + D
 + and so
( 2 x + 1 ) ( 2 x 2 + 1 ) 2 x + 1 2 x 2 + 1 | M1* | M1* | f(x) B C x + D
Re-writing , where f(x) is quadratic, as +
(2x+1)(2x2 +1) 2 x + 1 2 x 2 + 1
B C
or + and correct identity not involving fractions
2 x + 1 2 x 2 + 1
following through their partial fractions and quadratic f(x). | f(x) B C x + D
Re-writing , where f(x) is quadratic, as +
(2x+1)(2x2 +1) 2 x + 1 2 x 2 + 1
−6x2 −6x−3B(2x2 +1)+(Cx+D)(2x+1) | B C
or + and correct identity not involving fractions
2 x + 1 2 x 2 + 1
For example: Equating coefficients of x 2 : 2 B + 2 C = − 6
Let x = 0, gives B + D = − 3 | M1dep* | Equates coefficients or substitutes to find an equation involving only their
unknown(s).
A1 | Any (B = –1, C = –2, D = –2) one unknown correct from a correct partial
fraction form (so must have had a correct f(x)).
1 2 x + 2
3 − −
2 x + 1 2 x 2 + 1 | A1 | All unknowns correct – condone stating the correct form of the partial
fractions anywhere together with all four unknowns correctly stated without
necessarily bringing both parts together as a single expression at the end. So
B Cx+D
must see correct partial fraction expression or A+ + stated
2x+1 2x2 +1
and all correct values of A, B, C and D seen.
[5]
Equates coefficients or substitutes to find an equation involving only their
unknown(s).