Question

4u^2+8u find the zeros by splitting middle term and verify the relation between the zeros of this polynomial​

Answer 1

AnswEr :

☯ $\normalsize\sf\ Given \: Polynomial : 4u^2 + 8u$

$\underline{\bigstar\:\textsf{Zeroes \: of \: Polynomial:}}\\ \\ \\ \normalsize\ : \implies\sf\ 4u^2 + 8u = 0 \\ \qquad\footnotesize\star\sf\ Using \: Middle \: term \: factorization \\\\\\\normalsize\ : \implies\sf\ 4u(u + 2) = 0\\\\\\\normalsize\ : \implies\sf\ 4u = 0 \: \: or \: \: (u + 2) = 0\\\\\\\normalsize\ : \implies\sf\ u = \frac{0}{4} \: \: or \: \: 0 - 2\\\\\\\normalsize\ : \implies\sf\ u = 0 \: \: or \: \: -2\\\\\\\normalsize\ : \implies{\underline{\boxed{\sf \red{u = 0, -2}}}}\\\\\therefore\underline{\textsf{The \: Zeroes \: of \: Polynomial \: are \: }{\textbf{0, -2 }}}$

$\rule{100}2$

$\underline{\bigstar\:\sf{Relationship \: b/w \: zeroes \: \& \: Polynomial:}}\\\\\qquad\begin{aligned}\bf{\dag}\:\:\bf Sum \: of \: Zeroes\:\:\quad\end{aligned}\\\normalsize\dashrightarrow\sf\ \alpha + \beta = \frac{-b}{a}\\\\\\\normalsize\dashrightarrow\sf\ 0 + (-2) = \frac{\cancel{-8}}{\cancel{4}}\\\\\\\normalsize\dashrightarrow\sf\ -2 = -2\\\\\\\normalsize\dashrightarrow{\underline{\boxed{\sf \green{-2 = - 2}}}}\\\\\qquad\begin{aligned}\bf{\dag}\:\:\bf Product \: of \: Zeroes\:\:\quad\end{aligned}\\\normalsize\dashrightarrow\sf\ \alpha + \beta = \frac{c}{a}\\\\\\\normalsize\dashrightarrow\sf\ 0 \times\ (-2) = \frac{\cancel{0}}{\cancel{4}}\\\\\\\normalsize\dashrightarrow\sf\ 0 = 0\\\\\\\normalsize\dashrightarrow{\underline{\boxed{\sf \green{0 = 0}}}}\\\\\qquad\begin{aligned}\:\:\bf{Hence, Verified!!}\:\:\quad\end{aligned}$

Answer 2

$\Large\green{\underline{\underline{\bf{\green{Given}}}}}$

4u²+8u find the zeros by splitting middle term and verify the relation between the zeros of this polynomial

$\Large\red{\underline{\underline{\bf{\red{To\:find}}}}}$

Zeros of polynomial and verify the relationship between zeros of the polynomial

$\Large\purple{\underline{\underline{\bf{\purple{Solution}}}}}$

Let p(u) = 4u²+8u

p(u) = 0

so,

$\implies\sf 4u^2 + 8u = 0$

Take 4u as a common

$\implies\sf 4u(u+2)=0$

Either

=> 4u = 0

=> u = 0/4 = 0

or

=> u + 2 = 0

=> u = -2

So, the zeros of p(u) are 0 and -2

$\Large\green{\underline{\underline{\bf{\green{Verification}}}}}$

$\large{\boxed{\bf{\orange{Sum\:of\:zeros}}}}$

Add both the zeros

$\sf 0+(-2)=-2=\Large\frac{-(8)}{4}=\Large\frac{-coefficient\:of\:u}{coefficient\:of\:u^2}$

Now,

$\large{\boxed{\bf{\orange{Product\:of\:zeros}}}}$

Multiply both the zeros

$\sf 0\times{(-2)}=0=\Large\frac{0}{4}=\Large\frac{constant\:term}{coefficient\:of\:u^2}$