Introducing your new favourite teacher - Teachoo Black, at only βΉ83 per month
Ex 2.5, 1 (iii) - Chapter 2 Class 9 Polynomials (Term 2)
Last updated at Aug. 25, 2021 by Teachoo
Ex 2.5
Ex 2.5, 1 (i)
Ex 2.5, 1 (ii)
Ex 2.5, 1 (iii) Important You are here
Ex 2.5, 1 (iv)
Ex 2.5, 1 (v)
Ex 2.5, 2 (i)
Ex 2.5, 2 (ii) Important
Ex 2.5, 2 (iii)
Ex 2.5, 3 (i)
Ex 2.5, 3 (ii) Important
Ex 2.5, 3 (iii)
Ex 2.5, 4 (i)
Ex 2.5, 4 (ii)
Ex 2.5, 4 (iii) Important
Ex 2.5, 4 (iv)
Ex 2.5, 4 (v)
Ex 2.5, 4 (vi)
Ex 2.5, 5 (i)
Ex 2.5, 5 (ii) Important
Ex 2.5, 6 (i)
Ex 2.5, 6 (ii)
Ex 2.5, 6 (iii)
Ex 2.5, 6 (iv) Important
Ex 2.5, 7 (i)
Ex 2.5, 7 (ii)
Ex 2.5, 7 (iii) Important
Ex 2.5, 8 (i)
Ex 2.5, 8 (ii)
Ex 2.5, 8 (iii) Important
Ex 2.5, 8 (iv) Important
Ex 2.5, 8 (v)
Ex 2.5, 9 (i)
Ex 2.5, 9 (ii)
Ex 2.5, 10 (i) Important
Ex 2.5, 10 (ii)
Ex 2.5, 11
Ex 2.5,12 Important
Ex 2.5,13
Ex 2.5, 14 (i)
Ex 2.5, 14 (ii) Important
Ex 2.5, 15 (i)
Ex 2.5, 15 (ii) Important
Ex 2.5, 16 (i)
Ex 2.5, 16 (ii) Important
Transcript
Ex 2.5, 1 Use suitable identities to find the following products: (iii) (3x + 4) (3x - 5) (3x + 4) (3x β 5) = 3 (π₯+ 4/3) 3(π₯β 5/3) = (9) (π₯+ 4/3) (π₯β 5/3) Using (x + a) (x + b) = x2 + (a + b) x + ab, Where a = 4/3 , b = (β5)/3 = 9 [π₯2+(4/3+(β5)/3)π₯+(4/3)(β 5/3)] = 9 [π₯2 +((4 β 5))/3 π₯ β20/9] = 9 [π₯2 β 1/3 π₯ β20/9] = 9(π₯2) β9(1/3)π₯ β9(20/9) = 9x2 β 3x β 20
