find the G.C.D of the polynomials.1. 8(x³–²+x) and 28(x³+1)
Answers
Answer:
GiventwoAlgebraicexpressions:8(x3−x2+x)and[28(x3+1)]
We \:have ,Wehave,
\begin{gathered} 1) 8(x^{3} - x^{2} + x ) \\=\blue{ 2 \times 2} \times 2 x\blue{(x^{2} - x + 1 ) }\end{gathered}1)8(x3−x2+x)=2×2×2x(x2−x+1)
\begin{gathered} 2) 28(x^{3} + 1 ) \\= 2\times 2 \times 7 (x^{3} + 1^{3} ) \\= \blue{2\times 2 }\times 7 (x + 1 )\blue{(x^{2} - x + 1 )}\end{gathered}2)28(x3+1)=2×2×7(x3+13)=2×2×7(x+1)(x2−x+1)
\boxed {\pink {Since, a^{3} + b^{3} = (a+b)(a^{2} - ab + b^{2} }}Since,a3+b3=(a+b)(a2−ab+b2
\begin{gathered} Now , G.C.D \: of \: 8(x^{3} - x^{2} + x )\:and \\28(x^{3} + 1 ) \\\blue { = 2\times 2 \:(x^{2} - x + 1 )}\end{gathered}Now,G.C.Dof8(x3−x2+x)and28(x3+1)=2×2(x2−x+1)
\green { = 4(x^{2} - x + 1 )}=4(x2−x+1)
Therefore.,
\begin{gathered} \red{ G.C.D \: of \: 8(x^{3} - x^{2} + x )\:and }\\\red{28(x^{3} + 1 ) }\\\green { = 4(x^{2} - x + 1 )}\end{gathered}G.C.Dof8(x3−x2+x)and28(x3+1)=4(x2−x+1)