Question

find the quadratic polynomial in each case, with the given no. as its zeroes, 3,-5​

Answer 1

Let,

$\alpha = 3 \\ \\ \beta = - 5$

As we know that,

$=> P(x) = {x}^{2} - (\alpha + \beta)x + \alpha \beta$

$=> P(x) = {x}^{2} - [ 3 + (-5)]x + 3\times (-5)$

$=> P(x) = {x}^{2} + 2x - 15$

Answer 2

Let,

alpha = 3 and

beeta = -5

also,

alpha + beeta = 3-5 = -2 = -b/a

and alpha × beeta = 3×(-5) = -15 = c/a

we know that quardratic polynomial is ax2+bx+c

Now,

x2 - (alpha + beeta ) + alpha × beeta

=> x2- (-2)x + (-15)

=> x2+2x-15 Answer.......

hope it will help you.....