Question

Verify that the numbers 1/2,1,-2 are the zeroes of the cubic polynomial 2x³+x²-5x+2. Also,

verify the relationship between the zeroes and its coefficients.
​

Answer 1

Appropriate Question:

$\sf{Verify \: that \frac{1}{2} ,1, - 2 \: are \: the \: zeroes \: of \: the \: cubi \: polynomial \: {2x}^{3} + {x}^{2} - 5x + 2.} \\ \sf{Also \: verify \: the \: relationship \: between \: the \: zeroes \: and \: the \: coefficient.}$

Solution:

We Have, $\bf{p(x) = {2x}^{3} + {x}^{2} - 5x + 2 }$

Now,

$\longrightarrow \tt{p \bigg [ \frac{1}{2} \bigg] = 2 {\bigg[ \frac{1}{2} \bigg ]}^{3} } + {\bigg[ \frac{1}{2} \bigg ]}^{2} - 5 {\bigg[ \frac{1}{2} \bigg ]}^{2} + 2 \\ \\ \longrightarrow \tt{2 \times \frac{1}{8} + \frac{1}{4} - \frac{5}{2} + 2 } \\ \\ \longrightarrow \tt{ \frac{1}{4} + \frac{1}{4} - \frac{5}{2} + 2 } \\ \\ \longrightarrow \tt{ \frac{1 + 1 - 10 + 8}{4} } = 0$

$\tt{p(1) = 2( {1}^{3} ) + {1}^{2} - 5(1) + 2} \\ \tt{ = 2 + 1 - 5 + 2 }\\ \tt{ = 5 - 5 }\\ \tt{ = 0}$

$\tt{p( - 2) = 2 {( - 2)}^{2} + {( - 2)}^{2} - 5( - 2) + 2} \\ \tt{ = 2( - 8) + 4 + 10 + 2 }\\ \tt{ = - 16 + 16} \\ \tt{ = 0}$

$\tt{since \: p \bigg[ \frac{1}{2} \bigg ]=0 \: and \: p(-2)=0}$

$\tt{Hence, \frac{1}{2} ,1 \: and \: - 2 \: are \: zeroes \: of \: p(x) = {2x}^{3} + {x}^{2} - 5x + 2}.$

$\tt{let \: \alpha = \frac{1}{2} , \beta = 1, \gamma = - 2}$

Now,

$\leadsto\tt{\alpha + \beta + \gamma } \\ \tt{ \leadsto \frac{1}{2} + 1 - 2} \\ \tt{ \leadsto \frac{1}{2 } - 1 } \\ \tt{ \leadsto \frac{ - 1}{2} }$

$\: \: \: \: \: \: \tt{ \leadsto \frac{1}{2} = \frac{ - coefficiet \: of \: {x}^{2} }{coefficiet \: of \: {x}^{3} } }$

$\tt{\alpha \beta + \beta \gamma + \alpha \gamma = \frac{1}{2} (1) + (1)( - 2) + \frac{1}{2} ( - 2)} \\ \\ \tt{ = \frac{1}{2} - 2 - 1} \\ \\ \tt{= \frac{1}{2} - 3} \\ \\ \tt{ = \frac{1 - 6}{2} } \\ \\ \tt{ = \frac{ - 5}{2} = \frac{coefficiet \: of \: x}{coefficiet \: of \: {x}^{3} } }$

And,

$\longmapsto \tt{ \alpha \beta \gamma = \frac{1}{2} \times 1 \times ( - 2) } \\ \\ \tt{ \longmapsto - 1} \\ \\ \tt{ \longmapsto \frac{ - 2}{2} = \frac{constant \: term}{coefficiet \: of \: {x}^{3} } }$