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Dividing two polynomials
Ex 14.3, 4 (i)
Ex 14.3, 4 (v) Important
Ex 14.3, 4 (iv)
Ex 14.3, 4 (ii) Important
Ex 14.3, 4 (iii)
Ex 14.3, 3 (i)
Ex 14.3, 3 (ii)
Ex 14.3, 3 (iii) Important
Ex 14.3, 3 (iv)
Ex 14.3, 3 (v) Important
Ex 14.3, 5 (i)
Ex 14.3, 5 (ii) Important
Ex 14.3, 5 (iii)
Ex 14.3, 5 (iv) Important You are here
Ex 14.3, 5 (v)
Ex 14.3, 5 (vi)
Ex 14.3, 5 (vii) Important
Example 15 Important
Example 16
Last updated at Dec. 26, 2018 by Teachoo
Ex 14.3, 5 Factorise the expressions and divide them as directed. (iv) 4yz(π§^2 + 6z β 16) Γ· 2y(z + 8) We first factorise 4yz (π§^2+6π§β16) By middle term splitting, = 4yz (π§^2β2π§+8π§β16) = 4yz [(π§^2β2π§)+(8π§β16)] = 4yz [z (z β 2) + 8 (z β 2)] Taking (z β 2) common, = 4yz (z β 2) (z + 8) Splitting the middle term We need to find two numbers whose Sum = 6 Product = β16 So, we write 6z = β2z + 8z Dividing 4yz (π§^2+6π§β16)Γ·2π¦ (π§+8) = (4π¦π§ (π§^2 + 6π§ β 16))/(2π¦ (π§ + 8)) = (4π¦π§ (π§ β2) (π§ + 8))/(2π¦ (π§ + 8)) = 4/2 Γ π¦/π¦ Γ z Γ (z β 2) Γ ((π§ + 8))/((π§ + 8)) = 2 Γ z Γ (z β 2) = 2z (π β 2)