Question

9. If x = 4/3is a zero of the polynomial p(x) = 6x3– 11x2+ kx -20, then find the value of k.

Answer 1

Step-by-step explanation:

4/3 is zero of polynomial

BY REMAINDER THEOREM

P(4/3)=0

6*(4/3)³-11*2+K*4/3-20=0

6*64/27-22+4/3K-20=0

128/9-44+4/3k=0

4/3k=44-128/9

4/3k=396-128/9

4/3k=268/9

k=268/9*3/4

k=67/3ork=22.33333

Answer 2

$\huge{\bigstar}{ \green{ \mathfrak{ \underline{ \underline{Question}}}}}{\bigstar}$

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If $x = \displaystyle\frac{4}{3}$ is a zero of the polynomial $p(x) = 6x^3- 11x^2+ kx -20,$ then find the value of k

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$\huge{\bigstar}{ \green{ \mathfrak{ \underline{ \underline{Answer}}}}}{\bigstar}$

$\boxed{\red{\bold{Given:}}}$

$\displaystyle \frac{4}{3} \ is \ the \ zero \ of \ p(x)$

$\boxed{\red{\bold{Thus :}}}$

$\red{\implies}p \left( \displaystyle\frac{4}{3} \right) = 0$

${\red{\implies}}6 \left( \displaystyle \frac{4}{3} \right)^3 - 11 \left( \displaystyle\frac{4}{3} \right)^2 +k \left( \displaystyle\frac{4}{3} \right) -20 = 0$

${\red{\implies}}6 \times \displaystyle\frac{64}{27} - 11 \times \displaystyle \frac{16}{9} + \displaystyle\frac{4k}{3} -20 = 0$

$\green{\implies}\displaystyle \frac{128}{9} - \frac{176}{9} + \frac{4k}{3} = 20$

$\green{\implies}\displaystyle \frac{128 - 176 + 12k}{9} = 20$

$\green{\implies}-48 + 12k = 180$

$\green{\implies}12k = 180 + 48$

$\green{\implies}12k = 228$

$\green{\boxed{\implies{\boxed{k = 19}}}}$

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