Question

find the zeroes of the following polynomials by factorisation method and verify the relation
between the zeroes and the coefficients of the polynomials
t^3-2t^2-15t​

Answer 1

✪AnSwEr

${\tt{\red{\underline{\large{Given}}}}}$

• A polynomial
• t³-2t²-15t

${\sf{\green{\underline{\large{To\:Find}}}}}$

• Factors of the polynomial
• Relationship between cofficient

${\sf{\pink{\underline{\Large{Explanation}}}}}$

t³-2t²-15t=0

=>t(t²-2t-15)=0

=>t=0

=>(t²-2t-15)=0

• we have to spilt the middle term in such a way that the product become -15 and sum become -2t

=>t²-2t-15=0

=>t²-5t+3t-15=0

=>t(t-5)+3(t-5)=0

=>(t-5) (t+3)=0

=>t=5,-3

Let the zeroes of the polynomial be$\tt\alpha\beta{and}\gamma$

Then,

$\rightarrow\tt\alpha{+}\beta{+}\gamma{=}\frac{-b}{a}$

$\rightarrow\tt\alpha\beta+\alpha\gamma+\beta\gamma{=}\frac{c}{a}$

&

$\rightarrow\tt\alpha{\times}\beta{\times}\gamma{=}\frac{-d}{a}$

Here,

a=1

b=-2

C=-15

d=0

☞

$\rightarrow\tt\alpha{+}\beta{+}\gamma{=}\frac{2}{1}$

$\rightarrow\tt\alpha{+}\beta+\gamma{=}\dfrac{Cofficient\:of\:X^2}{Cofficient\:of\:x^3}=$

$\rightarrow\tt\alpha\beta+\alpha\gamma+\beta\gamma{=}\frac{-15}{0}$

$\rightarrow\tt\alpha{+}\beta+\gamma{=}\dfrac{Cofficient\:of\:X}{Cofficient\:of\:x^3}=$

&

$\rightarrow\tt\alpha{\times}\beta{\times}\gamma{=}\frac{0}{1}$

$\rightarrow\tt{\large\alpha{\times}\beta{=}\dfrac{Constant\:term}{Cofficient\:of\:x^3}}$

Hence,relation verified

Answer 2

$\bigstar$ Answer :

$\bold{t^3-2t^2-15t=0}\\\\t(t^2-2t-15)=0\\\\t(t^2+3t-5t-15)=0\\\\t(t(t+3)-5(t+3))=0\\\\t((t+3)(t-5))=0\\\\\bold{t(t+3)(t-5)=0}\\\\\implies\bold{ t=0\;,t=5\;,t=-3}$

$\bigstar$ Verification :

Let  α , β , γ be zeroes of cubic polynomial .

$\implies \alpha +\beta +\gamma=\frac{-b}{a}\\\\ \implies \alpha \beta +\alpha \gamma+\beta \gamma=\frac{c}{a}\\\\ \implies \alpha \beta \gamma=\frac{-d}{a}$

Now, compare given cubic polynomial , t³-2t²-15t with ax³+bx²+cx+d ,

we get ,

• a = 1 , b = -2 , c = -15 , d = 0 and
• α = 0 , β = 5 , γ = -3

$(1)\\\\\implies 0+5-3=\frac{-(-2)}{1}\\\\ \implies 2=2\\\\(2)\\\\\implies 0+0-15=\frac{-15}{1}\\\\ \implies -15=-15\\\\(3)\\\\\implies(0)(5)(-3)=\frac{-0}{1}\\\\ \implies0=0$