In the chap­ter on alge­braic iden­ti­ties you learned to expand (a+b)2(a + b)^{2}, (a−b)2(a - b)^{2} and (a+b)(a−b)(a + b)(a - b). Read back­wards, those same iden­ti­ties are fac­toris­ing tools. When­ever an expres­sion has the shape of the right-hand side of an iden­tity, you can write down its fac­tors at once. This les­son cov­ers the three square iden­ti­ties.

The three iden­ti­ties

a2−b2=(a+b)(a−b)a2+b2+2ab=(a+b)2a2+b2−2ab=(a−b)2\begin{aligned} a^{2} - b^{2} &= (a + b)(a - b) \\ a^{2} + b^{2} + 2ab &= (a + b)^{2} \\ a^{2} + b^{2} - 2ab &= (a - b)^{2} \end{aligned}

The skill is recog­nis­ing the shape. To do that, write each square term as a square: 49y249y^{2} as (7y)2(7y)^{2}, 3636 as 626^{2}, p4p^{4} as (p2)2(p^{2})^{2}. Once you can see what aa and bb are, the iden­tity does the rest.

Dif­fer­ence of two squares

An expres­sion of the form a2−b2a^{2} - b^{2}, one square minus another, is called a dif­fer­ence of two squares. It fac­torises as the sum times the dif­fer­ence: (a+b)(a−b)(a + b)(a - b).

Exam­ple 1

Fac­torise 49y2−3649y^{2} - 36.

49y2−36=(7y)2−62=(7y+6)(7y−6)\begin{aligned} 49y^{2} - 36 &= (7y)^{2} - 6^{2} \\ &= (7y + 6)(7y - 6) \end{aligned}

Here a=7ya = 7y and b=6b = 6. Check: (7y+6)(7y−6)=49y2−42y+42y−36=49y2−36(7y + 6)(7y - 6) = 49y^{2} - 42y + 42y - 36 = 49y^{2} - 36.

Exam­ple 2: using the iden­tity twice

Fac­torise p4−81p^{4} - 81.

p4−81=(p2)2−92=(p2+9)(p2−9)=(p2+9)(p2−32)=(p2+9)(p+3)(p−3)\begin{aligned} p^{4} - 81 &= (p^{2})^{2} - 9^{2} \\ &= (p^{2} + 9)(p^{2} - 9) \\ &= (p^{2} + 9)(p^{2} - 3^{2}) \\ &= (p^{2} + 9)(p + 3)(p - 3) \end{aligned}

After the first step, p2−9p^{2} - 9 is itself a dif­fer­ence of two squares, so it fac­torises again. Always look at each fac­tor and ask whether it can be bro­ken fur­ther. The fac­tor p2+9p^{2} + 9 is a sum of two squares, and it can­not be fac­torised this way.

Exam­ple 3: a com­mon fac­tor first

Fac­torise 3x2−273x^{2} - 27.

Nei­ther 3x23x^{2} nor 2727 is a per­fect square, but both have a fac­tor 33. Take it out first.

3x2−27=3(x2−9)=3(x+3)(x−3)\begin{aligned} 3x^{2} - 27 &= 3(x^{2} - 9) \\ &= 3(x + 3)(x - 3) \end{aligned}

Remem­ber. Before any­thing else, look for a com­mon fac­tor. Tak­ing it out first often reveals the shape of an iden­tity.

A use with num­bers

The iden­tity also makes some arith­metic easy:

1012−992=(101+99)(101−99)=200×2=400101^{2} - 99^{2} = (101 + 99)(101 - 99) = 200 \times 2 = 400

Per­fect squares

A three-term expres­sion of the form a2+b2+2aba^{2} + b^{2} + 2ab or a2+b2−2aba^{2} + b^{2} - 2ab is a per­fect square. To spot one, check three things: the first and last terms are squares, and the mid­dle term is exactly twice the prod­uct of their square roots.

Exam­ple 4

Fac­torise x2+8x+16x^{2} + 8x + 16.

The first term is x2x^{2} and the last is 16=4216 = 4^{2}. Twice their prod­uct is 2(x)(4)=8x2(x)(4) = 8x, which is exactly the mid­dle term.

x2+8x+16=x2+42+2(x)(4)=(x+4)2=(x+4)(x+4)\begin{aligned} x^{2} + 8x + 16 &= x^{2} + 4^{2} + 2(x)(4) \\ &= (x + 4)^{2} \\ &= (x + 4)(x + 4) \end{aligned}

Exam­ple 5

Fac­torise 25m2+30m+925m^{2} + 30m + 9.

25m2+30m+9=(5m)2+32+2(5m)(3)=(5m+3)2=(5m+3)(5m+3)\begin{aligned} 25m^{2} + 30m + 9 &= (5m)^{2} + 3^{2} + 2(5m)(3) \\ &= (5m + 3)^{2} = (5m + 3)(5m + 3) \end{aligned}

Exam­ple 6: a minus sign in the mid­dle

Fac­torise 4y2−12y+94y^{2} - 12y + 9.

The mid­dle term is neg­a­tive, so use a2+b2−2ab=(a−b)2a^{2} + b^{2} - 2ab = (a - b)^{2}.

4y2−12y+9=(2y)2+(3)2−2(2y)(3)=(2y−3)2=(2y−3)(2y−3)\begin{aligned} 4y^{2} - 12y + 9 &= (2y)^{2} + (3)^{2} - 2(2y)(3) \\ &= (2y - 3)^{2} \\ &= (2y - 3)(2y - 3) \end{aligned}

Note that both square terms, 4y24y^{2} and 99, are pos­i­tive. Only the mid­dle term car­ries the minus sign.

Com­bin­ing two iden­ti­ties

Exam­ple 7

Fac­torise a2−2ab+b2−c2a^{2} - 2ab + b^{2} - c^{2}.

The first three terms form a per­fect square, (a−b)2(a - b)^{2}. What is left is a dif­fer­ence of two squares.

a2−2ab+b2−c2=(a−b)2−c2=(a−b+c)(a−b−c)\begin{aligned} a^{2} - 2ab + b^{2} - c^{2} &= (a - b)^{2} - c^{2} \\ &= (a - b + c)(a - b - c) \end{aligned}

In the last step, (a−b)(a - b) plays the part of aa in a2−b2=(a+b)(a−b)a^{2} - b^{2} = (a + b)(a - b), and cc plays the part of bb. Treat a bracket as a sin­gle quan­tity and the iden­tity works just the same.

Prac­tice

Fac­torise.

  1. 81−16y281 - 16y^{2}
  2. 25a2−49b225a^{2} - 49b^{2}
  3. x4−16x^{4} - 16
  4. 2x2−502x^{2} - 50
  5. y2+10y+25y^{2} + 10y + 25
  6. 9p2−24pq+16q29p^{2} - 24pq + 16q^{2}
  7. 49m2+56m+1649m^{2} + 56m + 16
  8. x2+6x+9−y2x^{2} + 6x + 9 - y^{2}
  9. (2a+1)2−(a−3)2(2a + 1)^{2} - (a - 3)^{2}
  10. 4−x2+2xy−y24 - x^{2} + 2xy - y^{2}
  11. Use an iden­tity to work out 1052−952105^{2} - 95^{2}.

Answers

Show answers
  1. 92−(4y)2=(9+4y)(9−4y)9^{2} - (4y)^{2} = (9 + 4y)(9 - 4y)
  2. (5a)2−(7b)2=(5a+7b)(5a−7b)(5a)^{2} - (7b)^{2} = (5a + 7b)(5a - 7b)
  3. (x2+4)(x2−4)=(x2+4)(x+2)(x−2)(x^{2} + 4)(x^{2} - 4) = (x^{2} + 4)(x + 2)(x - 2)
  4. 2(x2−25)=2(x+5)(x−5)2(x^{2} - 25) = 2(x + 5)(x - 5)
  5. y2+52+2(y)(5)=(y+5)2y^{2} + 5^{2} + 2(y)(5) = (y + 5)^{2}
  6. (3p)2+(4q)2−2(3p)(4q)=(3p−4q)2(3p)^{2} + (4q)^{2} - 2(3p)(4q) = (3p - 4q)^{2}
  7. (7m)2+42+2(7m)(4)=(7m+4)2(7m)^{2} + 4^{2} + 2(7m)(4) = (7m + 4)^{2}
  8. (x+3)2−y2=(x+3+y)(x+3−y)(x + 3)^{2} - y^{2} = (x + 3 + y)(x + 3 - y)
  9. [(2a+1)+(a−3)][(2a+1)−(a−3)]=(3a−2)(a+4)[(2a + 1) + (a - 3)][(2a + 1) - (a - 3)] = (3a - 2)(a + 4)
  10. 4−(x2−2xy+y2)=22−(x−y)2=(2+x−y)(2−x+y)4 - (x^{2} - 2xy + y^{2}) = 2^{2} - (x - y)^{2} = (2 + x - y)(2 - x + y)
  11. (105+95)(105−95)=200×10=2000(105 + 95)(105 - 95) = 200 \times 10 = 2000