Ex 4.4, 1

Transcript

Ex 4.4, 1 (i) Fill in the blanks to complete the following identities: (i) 𝑠^2−11𝑠+24=( ____ ) ( ____ ) We need to factorise 𝒔^𝟐−𝟏𝟏𝒔+𝟐𝟒 𝑠^2−11𝑠+24 = 𝑠^2−𝟑𝐬−𝟖𝒔+24 = s(s−3)−8(s−3) = (𝐬−𝟑)(𝐬−𝟖) Splitting the middle term method We need to find two numbers whose Sum = –11 Product = 1 × 24 = 24 Since sum is negative but product is positive. Thus, both numbers are negative Thus, our fill in the blanks is 𝑠^2−11𝑠+24=(𝐬−𝟖)(𝐬−𝟑) Ex 4.4, 1 (ii) Fill in the blanks to complete the following identities: (ii) ( ____ (𝑥+1)=(3𝑥^2−4𝑥−7) We need to factorise 𝟑𝒙^𝟐−𝟒𝒙−𝟕 3𝑥^2−4𝑥−7 = 3𝑥^2−𝟕𝒙+𝟑𝒙−7 = 𝑥(3𝑥−7)+1(3𝑥−7) = (𝟑𝒙−𝟕)(𝒙+𝟏) Splitting the middle term method We need to find two numbers whose Sum = –4 Product = 3 × –7 = –21 Since both sum and product are negative. Thus, one number is positive, one is negative. And bigger is negative Thus, our fill in the blanks is (𝟑𝒙−𝟕)(𝑥+1)=3𝑥^2−4𝑥−7 Ex 4.4, 1 (iii) Fill in the blanks to complete the following identities: (iii) 10𝑥^2−11𝑥−6=(2𝑥−____" ) ( "____" + 2)" We need to factorise 𝟏𝟎𝒙^𝟐−𝟏𝟏𝒙−𝟔 10𝑥^2−11𝑥−6 = 10𝑥^2−𝟏𝟓𝒙+𝟒𝒙−6 = 5𝑥(2𝑥−3)+2(2𝑥−3) = (𝟐𝒙−𝟑)(𝟓𝒙+𝟐) Splitting the middle term method We need to find two numbers whose Sum = –11 Product = 10 × –6 = –60 Since both sum and product are negative. Thus, one number is positive, one is negative. And bigger is negative Thus, our fill in the blanks is 10𝑥^2−11𝑥−6=(𝟐𝒙−𝟑)(𝟓𝒙+𝟐) Ex 4.4, 1 (iv) Fill in the blanks to complete the following identities: (iv) 6𝑥^2+7𝑥+2=(" "____" ) ( "____" )" We need to factorise 𝟔𝒙^𝟐+𝟕𝒙+𝟐 6𝑥^2+7𝑥+2 = 6𝑥^2+𝟒𝒙+𝟑𝒙+2 = 2𝑥(3𝑥+2)+1(3𝑥+2) = (𝟐𝒙+𝟏)(𝟑𝒙+𝟐) Splitting the middle term method We need to find two numbers whose Sum = 7 Product = 6 × 2 = 12 Since sum and product both are positive, both numbers are positive Thus, our fill in the blanks is 6𝑥^2+7𝑥+2=(𝟑𝒙+𝟐)(𝟐𝒙+𝟏)