Question

Find a quadratic polynomial, the sum and product of whose zeroes are

-6 and 8, respectively. ​

Answer 1

${\huge{\red{\sf{Given}}}}\begin{cases}\leadsto Sum\:of\:zeroes\:is\:-6\\\leadsto Product\:of\:zeroes\:is\:8\end{cases}$

${\huge{\red{\sf{To\:Find}}}}\begin{cases}\leadsto To\:find\:the\:quadratic\:polynomial\end{cases}$

$\huge\red{\underline{\bf{\green{Answer}}}}$

We have,

• Sum of zeroes as -6.
• Product of zero as 8.

So, we know,

$\large\red{\boxed{\sf{\purple{p(x)=k[x^{2}-(\alpha+\beta)x+\alpha\beta]}}}}$

where,

• $\alpha$&$\beta$ are zeroes.
• p(x) is required polynomial.
• k is a non-zero constant.

______________________________________

Using above formula,

$\implies p(x) =k[x^{2}-(-6)x+8]$

$\implies p(x) =k[x^{2}+6x+8]$

${\underline{\boxed{\red{\bf{\mapsto p(x) =k[x^{2}+6x+8]}}}}}$

Answer 2

Step-by-step explanation:

$\alpha + \beta = - 6$

$\alpha \beta = 8 \\ p(x) = x ^{2} - x( \alpha + \beta ) + ( \alpha \beta )$

$p(x) = x ^{2} - x( - 6) + (8)$

$p(x) = x ^{2} + 6x + 8$