Question

if alfa & bita are zero of polynomial 3x^2-7x-6 find polynomials whose zero are 2 alfa+ 3bita & 3 alfa + 2 bita​

Answer 1

Solution :

We have quadratic polynomial p(x) = 3x² - 7x - 6 & zero of the polynomial p(x) = 0

Now;

$\longrightarrow\sf{3x^{2} -7x - 6=0}\\\\\longrightarrow\sf{3x^{2} -9x + 2x -6=0}\\\\\longrightarrow\sf{3x(x-3) +2(x-3) = 0}\\\\\longrightarrow\sf{(x-3)(3x+2) = 0}\\\\\longrightarrow\sf{x-3=0\:\:Or\:\:3x+2=0}\\\\\longrightarrow\sf{x=3\:\:Or\:\:3x=-2}\\\\\longrightarrow\bf{x=3\:\:Or\:\:x=-2/3}$

∴ α = 3 & β = -2/3 are the zeroes of the given polynomial .

Now;

We have another zeroes 2α + 3β & 3α + 2β, we get;

$\underline{\mathcal{SUM\:OF\:THE\:ZEROES\::}}$

$\mapsto\sf{\alpha+\beta =\dfrac{-b}{a} =\bigg\lgroup \dfrac{Coefficient\:of\:x}{Coefficient\:of\:x^{2}} \bigg\rgroup }\\\\\mapsto\sf{(2\alpha +3\beta ) + (3\alpha +2\beta )}\\\\\mapsto\sf{2\alpha+ 3\beta +3\alpha +2\beta}\\\\\mapsto\sf{2\alpha + 3\alpha + 3\beta +2\beta }\\\\\mapsto\sf{5\alpha + 5 \beta }\\\\\mapsto\sf{5(\alpha +\beta )}\\\\\mapsto\sf{5\bigg[3+\bigg(-\dfrac{2}{3} \bigg)\bigg]}\\\\\\\mapsto\sf{5\bigg[3 -\dfrac{2}{3}\bigg]}\\\\\\\mapsto\sf{5\bigg[\dfrac{9-2}{3}\bigg]}$

$\mapsto\sf{5\times \dfrac{7}{3} }\\\\\\\mapsto\bf{ \dfrac{35}{3} }$

$\underline{\mathcal{PRODUCT\:OF\:THE\:ZEROES\::}}$

$\mapsto\sf{\alpha\times \beta =\dfrac{c}{a} =\bigg\lgroup \dfrac{Constant\:term}{Coefficient\:of\:x^{2}} \bigg\rgroup }\\\\\mapsto\sf{(2\alpha +3\beta ) \times (3\alpha +2\beta )}\\\\\mapsto\sf{2\alpha (3\alpha + 2\beta ) + 3\beta ( 3\alpha +2\beta)}\\\\\mapsto\sf{6\alpha^{2} +4\alpha \beta +9\alpha \beta + 6\beta ^{2}}\\ \\\mapsto\sf{6(\alpha ^{2} + \beta^{2} ) +13\alpha\beta }\\\\\mapsto\sf{6(\alpha +\beta)^{2} -2\alpha \beta + 13 \alpha \beta }$

$\mapsto\sf{6(\alpha +\beta)^{2} + 11 \alpha \beta }\\\\\mapsto\sf{6\bigg(\dfrac{7}{3} \bigg)^{2} + 11\bigg[3\times \bigg(\dfrac{-2}{3}\bigg) \bigg]}\\\\\\\mapsto\sf{\cancel{6}\times \dfrac{49}{\cancel{9}} + 11 \cancel{\bigg(\dfrac{-6}{3}\bigg)}}\\\\\\\mapsto\sf{2\times \dfrac{49}{3} +11 (-2) }\\\\\\\mapsto\sf{ \dfrac{98}{3} + (-22) }\\\\\\\mapsto\sf{ \dfrac{98}{3} -22 }\\\\\\\mapsto\sf{ \dfrac{98-66}{3} }\\\\\\\mapsto\bf{ \dfrac{32}{3} }$

Thus;

The required polynomials are;

$\longrightarrow\sf{x^{2} - (sum\:of\:zeroes)x + (product\:of\:zeroes)}\\\\\longrightarrow\sf{x^{2} - \bigg(\dfrac{35}{3} \bigg)x+\bigg(\dfrac{32}{3} \bigg)}\\\\\longrightarrow\bf{3x^{2} -35x + 32}$