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Get live Maths 1-on-1 Classs - Class 6 to 12
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Last updated at March 22, 2023 by Teachoo
Example 5 Simplify the expressions and evaluate them as directed: (ii) 3π¦(2π¦β7)β3(π¦β4)β63 for π¦=β2 3π¦(2π¦β7)β3(π¦β4)β63 =3π¦Γ2π¦β3π¦Γ7β3Γπ¦+3Γ4β63 =6π¦^2β21π¦β3π¦+12β63 =6π¦^2β24π¦β51 For y = βπ Putting π¦=β2 in expression 6π¦^2β24π¦β51 = 6(β2)^2β24(β2)β51 =6Γ4+24Γ2β51 = 24+48β51 = 72β51 = ππ