( 3 z + 2 k ) 2 = ( 3 z ) 2 + 2 ⋅ 3 z ⋅ 2 k + ( 2 k ) 2 = 9 z 2 + 12 k z + 4 k 2
You can solve this by the 1st identity which is (a+b)=a^2+b^2+2ab (3z)^2+(2k)^2+2 3z 2k =9z^2+4k^2+12kz
To solve ( 3 z + 2 k ) 2 , use the binomial expansion formula to get 9 z 2 + 12 z k + 4 k 2 . This involves squaring each term and adding them together, including the cross term. The final expanded form is 9 z 2 + 12 z k + 4 k 2 .
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[tex] = \frac{5 \times 30}{15} [/tex][tex] = \frac{150}{15} [/tex]= 10 hari[tex] \pink{\mathcal{M \it 01}} [/tex]