Question

find the zeroes of the polynomial polynomial x3_5x2_2x+24 lf it is given that the product of two zeroes is 12​

Answer 1

$\huge{\underline{\purple{\rm{Solution:}}}}$

$\large{\underline{\red{\rm{let:}}}}$

• $\sf{The\:zeroes\:of\: polynomial\be\:\alpha,\beta,\gamma}$

$\small{\underline{\green{\rm{Given:}}}}$

• $\rm{f(x)=x^3-5x^2-2x+24}$
• $\rm{The\: product\:2\: zeroes=12}$

$\small{\underline{\orange{\rm{To\: Find:}}}}$

• $\sf{The\: zeroes\:of\:polynomial=?}$

$\sf\large{Formula\: used\:to\: calculate\: zeroes}$

$\small{\underline{\green{\rm{Sum\:of\:zeroes\:of\:f(x):}}}}$

$\sf{\implies \alpha+\beta+\gamma=\frac{-\:cofficient\:of\:x^2}{cofficient\:of\:x^3}}$

$\small{\underline{\red{\rm{Product\:of\:zeroes\:of\:f(x):}}}}$

$\sf{\implies \alpha\beta\gamma=\frac{-constant\:term}{cofficient\:of\:x^3}}$

$\small{\underline{\orange{\rm{According\:to\: above\: information:}}}}$

$\rm{The\: product\:2\: zeroes=12}$

$\rm{\implies \alpha*\beta=12}$

$\rm\purple{\implies \alpha\beta=12-------(i)}$

$\rm{\implies \alpha+\beta+\gamma=\dfrac{-(-5)}{1}}$

$\rm\purple{\implies \alpha+\beta+\gamma=5-----(ii)}$

$\rm{\implies \alpha\beta\gamma=\dfrac{-24}{1}}$

$\rm\purple{\implies \alpha\beta\gamma=-24------(iii)}$

• $\sf{Putting\:the\: value\:of\:\alpha\beta=12\:in\:eq\:iii}$

$\rm{\implies \alpha\beta\gamma=-24}$

$\rm{\implies 12\gamma=-24}$

$\rm\red{\implies \gamma=-2}$

• $\sf{Putting\:the\: value\:of\:\gamma=-2\:in\:eq\:ii}$

$\rm{\implies \alpha+\beta+\gamma=5}$

$\rm{\implies \alpha+\beta-2=5}$

$\rm{\implies \alpha+\beta=5+2}$

$\rm\purple{\implies \alpha+\beta=7------(iv)}$

$\rm{\implies \alpha=7-\beta}$

• $\sf{squaring\:on\: both\: sides}$

$\rm{\implies (\alpha+\beta)^2=7^2}$

$\rm{\implies \alpha^2+\beta^2+2\alpha\beta=7^2}$

• $\sf{Putting\:the\: value\:of\:\alpha\beta=12\:in\:eq\:iv}$

$\rm{\implies \alpha^2+\beta^2+2*12=7^2}$

$\rm{\implies \alpha^2+\beta^2+24=49}$

$\rm{\implies \alpha^2+\beta^2=49-24}$

$\rm\purple{\implies \alpha^2+\beta^2=25-------(v)}$

• $\sf{putting\:\alpha=7-\beta\:here}$

$\rm{\implies \alpha^2+\beta^2=25}$

$\rm{\implies (7-\beta)^2+\beta^2=25}$

$\rm{\implies 49-14\beta+\beta^2+\beta^2=25}$

$\rm{\implies 2\beta^2-14\beta+49-25=0}$

$\rm{\implies 2\beta^2-14\beta+24=0}$

• $\sf{dividing\:by\:2\:on\:side\:here}$

$\rm{\implies \beta^2-7\beta+12=0}$

• $\sf{Splitting\: middle\: term\: here}$

$\rm{\implies \beta^2-3\beta-4\beta+12=0}$

$\rm{\implies \beta(\beta-3)-4(\beta-3)=0}$

$\rm{\implies (\beta-4)(\beta-3)=0}$

$\rm\red{\implies \beta=4\:and\:3}$

• $\sf{both\: value\:are\: positive\:so\:we\:take\: anyo}$
• $\sf{The\: value\:of\:\beta=3\:here}$

$\rm{\implies \alpha+\beta=7}$

$\rm{\implies \alpha+3=7}$

$\rm{\implies \alpha=7-3}$

$\rm\red{\implies \alpha=4}$

Hence,

$\sf\orange{\implies The\: zeroes\:of\: polynomial=4,\:3,\:and\:-2}$