how we take common (k+1) from above eq.
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I hope that this helps
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=>4k3+18k2+23k+9/3
=>4k3+18k2+14k+9k+9/3
=>4k3+4k2+14k2+14k+9k+9/3
=>4k2(k+1)+14k(k+1)+9(k+1)/3
Now (k+1) is common factor..
then
=>(k+1)(4k2+14k+9)/3
hope it will help you..
=>4k3+18k2+14k+9k+9/3
=>4k3+4k2+14k2+14k+9k+9/3
=>4k2(k+1)+14k(k+1)+9(k+1)/3
Now (k+1) is common factor..
then
=>(k+1)(4k2+14k+9)/3
hope it will help you..
vvbnmittalp8bd27:
but its complicated
yes
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