How do you factor the trinomial 18a^2+48a+3218a2+48a+32?

1 Answer
Nov 18, 2015

2(3a+4)^22(3a+4)2

Explanation:

18a^2 + 48a + 3218a2+48a+32

  • First reduce it to the smallest terms
  • They are all divisible by 2

(18a^2)/2 + (48a)/2 + 32/218a22+48a2+322

9a^2 + 24a + 169a2+24a+16

16(last term) x 9(co-efficient of a^2a2) = 144

  • Think of two numbers that
    rArr When you multiply them = 144
    rArr When you add them = 24 (co-efficient of aa)

  • 12 and 12
    rArr 12 x 12 = 144
    rArr 12 + 12 = 24

  • Replace the middle number with 12 and 12
    9a^2 + 12a + 12a + 169a2+12a+12a+16

  • Split the expression
    9a^2 + 12a----- | ----- + 12a + 169a2+12a+12a+16

  • Factorise the two
    3a(3a + 4) + 4(3a +4)3a(3a+4)+4(3a+4)

  • Common factors so group
    (3a + 4)(3a +4)(3a+4)(3a+4)

rArr (3a+4)^2(3a+4)2

  • Remember we divided by 2
    rArr 2(3a+4)^22(3a+4)2