Multiplication of two polynomials [tex]\left(3 x^{2}+4 x+4\right) \text { and }(x-4) \text { is } 3 x^{3}-8 x^{2}-12 x-16[/tex]
Solution:
Need to multiply following polynomial
[tex]\left(3 x^{2}+4 x+4\right)(x-4)[/tex]
[tex]\text { Lets for sake of simplicity }\left(3 x^{2}+4 x+4\right)=\mathrm{A}[/tex]
[tex]\text { So }\left(3 x^{2}+4 x+4\right) \times(x-4)=\mathrm{A} \times(x-4)=\mathrm{A} \times(x)-\mathrm{A} \times 4[/tex]
[tex]\text { Now again substitute value of } \mathrm{A} \text { as }\left(3 x^{2}+4 x+4\right)[/tex]
[tex]=>\left(3 x^{2}+4 x+4\right) \times(x-4)=\left[\left(3 x^{2}+4 x+4\right) \times(x)\right]-\left[\left(3 x^{2}+4 x+4\right) \times 4\right][/tex]
On opening the internal brackets we get,
[tex]\begin{array}{l}{\Rightarrow\left(3 x^{2}+4 x+4\right) \times (x-4)=\left[3 x^{3}+4 x^{2}+4 x\right]-\left[12 x^{2}+16 x+16\right]} \\\\ {\Rightarrow\left(3 x^{2}+4 x+4\right) \times (x-4)=3 x^{3}+4 x^{2}+4 x-12 x^{2}-16 x-16}\end{array}[/tex]
On bringing similar terms together, we get
[tex]\begin{array}{l}{=>\left(3 x^{2}+4 x+4\right) \times (x-4)=3 x^{3}+4 x^{2}-12 x^{2}+4 x-16 x-16} \\ {=>\left(3 x^{2}+4 x+4\right) \times (x-4)=3 x^{3}-8 x^{2}-12x-16}\end{array}[/tex]
Hence the product is found out
