When x3+2x2 + kx +3 is …
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Sia ? 6 years, 6 months ago
Let p(x) ={tex} x^3 + 2x^2 + kx + 3{/tex}
Now, x - 3 = 0
{tex}\Rightarrow{/tex} x = 3
By the remainder theorem, we know that when p(x) is divided by (x - 3), the remainder is p(3).
Now, p(3) = (3)3 + 2(3)2 + k(3) + 3
= 27 + 2(9) + 3k + 3
= 30 + 18 + 3k
= 48 + 3k
But, remainder = 21
{tex}\Rightarrow{/tex} 48 + 3k = 21
{tex}\Rightarrow{/tex} 3k = 21 - 48
{tex}\Rightarrow{/tex} 3k = -27
{tex}\Rightarrow{/tex} k = {tex}\frac{-27}3{/tex}
{tex}\Rightarrow{/tex} k = -9
So, the value of k is -9.
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