p(x) =kx^3 + 3x^2 - 3 and q(x) =2x^3 - 5x + k, when divided by x-4 leave the same remainder in each case, find the value of k.
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p(x) =kx^3 + 3x^2 - 3 and q(x) =2x^3 - 5x + k, when divided by x-4 leave the same remainder in each case, find the value of k.
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[tex]\large{\red{\underline{\underline{\bold{Given:-}}}}}[/tex]
[tex]\large{\blue{\underline{\underline{\bold{To Find:-}}}}}[/tex]
[tex]\large{\green{\underline{\underline{\bold{Solution:-}}}}}[/tex]
[tex]\rightarrow[/tex] {As we know that we should use remainder theorem.Remainder theorem says that when x-4 then the value of x is 4.So substitute x in place of p(x) and f(x)}
[tex]\rightarrow[/tex] p(x)= kx³+3x²-3 , q(x)= 2x³-5x+k
[tex]\rightarrow[/tex] p(4)= k(4)³+3(4)²-3,q(4)=2(4)³-5(4)+k
[tex]\rightarrow[/tex] p(4)= 64k+48-3 , q(4)= 128-20+k
[tex]\rightarrow[/tex] p(4)= 64k+45 , q(4)= 108+k
[tex]\longrightarrow[/tex] 64k+45..............(i)
[tex]\longrightarrow[/tex] k+108.................(ii)
[tex]\rightarrow[/tex] 64k+45= k+108
[tex]\rightarrow[/tex] 64k-k= 108-45
[tex]\rightarrow[/tex] 63k= 63
[tex]\rightarrow[/tex] k= 1