Question

Find the zeroes of the polynomial f(x)=x3
-5x2
-2x+24 if it is given that the product of its two
zeroes is 12.

Answer 1

Solution :

We have cubic polynomial p(x) = x³ - 5x² - 2x + 24

As we know that given polynomial compared with ax³ + bx² + cx + d;

• a = 1
• b = -5
• c = -2
• d = 24

$\sf{Let\:\alpha \:,\:\beta\: \&\:\gamma\:are\:the\:zereos\:of\:the\:polynomial.}$

A/q

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

$\longrightarrow\sf{\alpha +\beta +\gamma=\dfrac{-b}{a}=-\bigg\lgroup\dfrac{Coefficient\:of\:x^{2} }{Coefficient\:of\:x^{3} }\bigg\rgroup }}\\\\\\\longrightarrow\sf{\alpha +\beta +\gamma=\dfrac{-(-5)}{1} }\\\\\\\longrightarrow\bf{\alpha +\beta +\gamma=5......................(1)}$

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

$\longrightarrow\sf{\alpha \beta \gamma=\dfrac{-d}{a}=-\bigg\lgroup\dfrac{Constant\:term }{Coefficient\:of\:x^{3} }\bigg\rgroup }}\\\\\\\longrightarrow\sf{\alpha \beta \gamma=\dfrac{-24}{1} }\\\\\\\longrightarrow\bf{12\gamma=-24\:\:\:[\therefore \:\alpha \beta =12]}\\\\\\\longrightarrow\sf{\gamma=\cancel{\dfrac{-24}{12} }}\\\\\\\longrightarrow\bf{\gamma=-2}$

Putting the value of gamma in equation (1),we get;

$\longrightarrow\sf{\alpha +\beta +(-2)=5}\\\\\longrightarrow\sf{\alpha +\beta -2=5}\\\\\longrightarrow\sf{\alpha +\beta =5+2}\\\\\longrightarrow\sf{\alpha +\beta =7}\\\\\longrightarrow\bf{\alpha =7-\beta ........................(2)}$

Putting the value of alpha in product of the two zeroes :

$\longrightarrow\sf{\alpha \beta =12}\\\\\longrightarrow\sf{(7-\beta )\beta =12\:\:[from\:(2)]}\\\\\longrightarrow\sf{7\beta -\beta ^{2} =12}\\\\\longrightarrow\sf{\beta ^{2} -7\beta +12=0}\\\\\longrightarrow\sf{\beta ^{2} -4\beta -3\beta +12=0}\\\\\longrightarrow\sf{\beta (\beta -4)-3(\beta -4)=0}\\\\\longrightarrow\sf{(\beta -4)(\beta -3)=0}\\\\\longrightarrow\sf{\beta -4=0\:\:\:Or\:\:\:\beta -3=0}\\\\\longrightarrow\bf{\beta =4\:\:\:Or\:\:\:\beta =3}$

Putting the value of both beta in equation (2),we get;

$\longrightarrow\sf{\alpha =7-4}\\\\\longrightarrow\bf{\alpha =3}$

&

$\longrightarrow\sf{\alpha =7-3}\\\\\longrightarrow\bf{\alpha =4}$

Thus;

$\boxed{\sf{The\:zeroes\:are\:\alpha =3,4 \:\:\& \:\:\beta=4,3\:\: \&\:\:\gamma=-2}}}}$