Question

Verify that the numbers given alongside the cubic polynomials below are their zeros. Also Verify the relationship between the Zeros and the coefficients.

(i)
${x}^{3} - 2x - 5x + 6 \\ zeros = - 2 \: \: \: \: \: \: \: \: 1 \: \: \: \: \: and \: \: 3$
(ii)
${2x}^{3} + 7x + 2x - 3 \\ zeros \: = - 3 \: \: \: \: \: \: \: \: \: \: \: - 1\: \: \: \: \: \: \: and \: \frac{1}{2}$
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Answer 1

$</p><p>\begin{gathered}if \: \alpha \: and \: beta \: are \: the \: zeroes \: of \: the \: polynomial \\ \: f(x) = {4x}^{2} - 5x + 1 \: find \: a \: quadratic \: \\ polynomial \: whose \: zeros \: are \: \\ \frac{ \alpha ^{2} }{ \beta } and \: \frac{{ \beta }^{2}}{ \alpha } \end{gathered}ifαandbetaarethezeroesofthepolynomialf(x)=4x2−5x+1findaquadraticpolynomialwhosezerosareβα2andαβ2</p><p>$

$\begin{gathered}\bold{\pink{Solution}}\longrightarrow \\ f(x) = 4 {x}^{2} - 5x + 1 \\ = 4 {x}^{2} - (4 + 1)x + 1 \\ = 4 {x}^{2} - 4x - x + 1 \\ = 4x(x - 1) - 1(x - 1) \\ = (x - 1) \: (4x - 1) \\ if \: x - 1 = 0 \: \: \: = > x = 1 \\ if \: 4x - 1 = 0 = > x = \dfrac{1}{4} \\ so \\ \alpha = 1 \: and \: \beta = \dfrac{1}{4} \\ \dfrac{ { \alpha }^{2} }{ \beta } = \dfrac{ {(1)}^{2} }{ \dfrac{1}{4} } \\ \dfrac{ { \alpha }^{2} }{ \beta } = 4 \\ \dfrac{ { \beta }^{2} }{ \alpha } = \dfrac{( \dfrac{ {1} }{4}) ^{2} }{1} \\ \dfrac{ { \beta }^{2} }{ \alpha } = \dfrac{1}{16} \\ now\end{gathered}$

$\begin{gathered}sum \: of \: zeroes \: = \dfrac{ { \alpha }^{2} }{ \beta } + \dfrac{ { \beta }^{2} }{ \alpha } \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = 4 + \dfrac{1}{16} \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \dfrac{65}{16} \end{gathered}$

$\begin{gathered}product \: of \: zeroes = \dfrac{ { \alpha }^{2} }{ \beta } \: \: \: \dfrac{ { \beta }^{2} }{ \alpha } \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \alpha \beta \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \dfrac{1}{4} \end{gathered}$

$\begin{gathered}required \: polynomial \: is \\ k( {x}^{2} - (sum \: of \: zeroes)x + product \: of \: zeroes \: ) \\ = k( {x}^{2} - ( \dfrac{65}{16} )x + \dfrac{1}{4} ) \\ = k( \dfrac{16 {x}^{2} - 65x + 4}{16} ) \\ taking \: k \: = 16 \: we \: get \\ = 16( \dfrac{16 {x}^{2} - 65x + 4}{16} ) \\ = \pink{16 {x}^{2} - 65x + 4}\end{gathered}$

Answer 2

Answer:

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