Math, asked by savita9561, 8 months ago

find the product of binomial(m^2-5) (6-mn^2)​

Answers

Answered by dipanshuashoka90
3

To multiply a binomial, four multiplications must take place. The first two terms need to be multiplied by the other two terms as follows (the order of multiplication is not important as long as the correct terms are being multiplied):

To multiply a binomial, four multiplications must take place. The first two terms need to be multiplied by the other two terms as follows (the order of multiplication is not important as long as the correct terms are being multiplied):\begin{gathered}(m^{2}n-5)(6-mn^{2})\\=6m^2n-m^3n^3-30+5mn^2\end{gathered}

To multiply a binomial, four multiplications must take place. The first two terms need to be multiplied by the other two terms as follows (the order of multiplication is not important as long as the correct terms are being multiplied):\begin{gathered}(m^{2}n-5)(6-mn^{2})\\=6m^2n-m^3n^3-30+5mn^2\end{gathered} (m

To multiply a binomial, four multiplications must take place. The first two terms need to be multiplied by the other two terms as follows (the order of multiplication is not important as long as the correct terms are being multiplied):\begin{gathered}(m^{2}n-5)(6-mn^{2})\\=6m^2n-m^3n^3-30+5mn^2\end{gathered} (m 2

To multiply a binomial, four multiplications must take place. The first two terms need to be multiplied by the other two terms as follows (the order of multiplication is not important as long as the correct terms are being multiplied):\begin{gathered}(m^{2}n-5)(6-mn^{2})\\=6m^2n-m^3n^3-30+5mn^2\end{gathered} (m 2 n−5)(6−mn

To multiply a binomial, four multiplications must take place. The first two terms need to be multiplied by the other two terms as follows (the order of multiplication is not important as long as the correct terms are being multiplied):\begin{gathered}(m^{2}n-5)(6-mn^{2})\\=6m^2n-m^3n^3-30+5mn^2\end{gathered} (m 2 n−5)(6−mn 2

To multiply a binomial, four multiplications must take place. The first two terms need to be multiplied by the other two terms as follows (the order of multiplication is not important as long as the correct terms are being multiplied):\begin{gathered}(m^{2}n-5)(6-mn^{2})\\=6m^2n-m^3n^3-30+5mn^2\end{gathered} (m 2 n−5)(6−mn 2 )

To multiply a binomial, four multiplications must take place. The first two terms need to be multiplied by the other two terms as follows (the order of multiplication is not important as long as the correct terms are being multiplied):\begin{gathered}(m^{2}n-5)(6-mn^{2})\\=6m^2n-m^3n^3-30+5mn^2\end{gathered} (m 2 n−5)(6−mn 2 )=6m

To multiply a binomial, four multiplications must take place. The first two terms need to be multiplied by the other two terms as follows (the order of multiplication is not important as long as the correct terms are being multiplied):\begin{gathered}(m^{2}n-5)(6-mn^{2})\\=6m^2n-m^3n^3-30+5mn^2\end{gathered} (m 2 n−5)(6−mn 2 )=6m 2

To multiply a binomial, four multiplications must take place. The first two terms need to be multiplied by the other two terms as follows (the order of multiplication is not important as long as the correct terms are being multiplied):\begin{gathered}(m^{2}n-5)(6-mn^{2})\\=6m^2n-m^3n^3-30+5mn^2\end{gathered} (m 2 n−5)(6−mn 2 )=6m 2 n−m

To multiply a binomial, four multiplications must take place. The first two terms need to be multiplied by the other two terms as follows (the order of multiplication is not important as long as the correct terms are being multiplied):\begin{gathered}(m^{2}n-5)(6-mn^{2})\\=6m^2n-m^3n^3-30+5mn^2\end{gathered} (m 2 n−5)(6−mn 2 )=6m 2 n−m 3

To multiply a binomial, four multiplications must take place. The first two terms need to be multiplied by the other two terms as follows (the order of multiplication is not important as long as the correct terms are being multiplied):\begin{gathered}(m^{2}n-5)(6-mn^{2})\\=6m^2n-m^3n^3-30+5mn^2\end{gathered} (m 2 n−5)(6−mn 2 )=6m 2 n−m 3 n

To multiply a binomial, four multiplications must take place. The first two terms need to be multiplied by the other two terms as follows (the order of multiplication is not important as long as the correct terms are being multiplied):\begin{gathered}(m^{2}n-5)(6-mn^{2})\\=6m^2n-m^3n^3-30+5mn^2\end{gathered} (m 2 n−5)(6−mn 2 )=6m 2 n−m 3 n 3

To multiply a binomial, four multiplications must take place. The first two terms need to be multiplied by the other two terms as follows (the order of multiplication is not important as long as the correct terms are being multiplied):\begin{gathered}(m^{2}n-5)(6-mn^{2})\\=6m^2n-m^3n^3-30+5mn^2\end{gathered} (m 2 n−5)(6−mn 2 )=6m 2 n−m 3 n 3 −30+5mn

To multiply a binomial, four multiplications must take place. The first two terms need to be multiplied by the other two terms as follows (the order of multiplication is not important as long as the correct terms are being multiplied):\begin{gathered}(m^{2}n-5)(6-mn^{2})\\=6m^2n-m^3n^3-30+5mn^2\end{gathered} (m 2 n−5)(6−mn 2 )=6m 2 n−m 3 n 3 −30+5mn 2

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