Question

Find the quadratic polynomial sum of whose zeros is 8 and their product is 12 find the zeros of the polynomial

Answer 1

Answer:

• The required Polynomial is x² - 8x + 12 and its zeros are 6 & 2.

Step-by-step explanation:

We have been given that Sum of zeros and Product of Zeros are 8 and 12 respectively

• Let zeros be a and ß for the required Polynomial.

So We have:

• Sum of Zeros ( a + ß ) = 8
• Sum of Zeros ( a + ß ) = 8 Product of Zeros ( aß) = 12

Now, We have to find Polynomial with the given zeros.

• General Formula of Quadratic Polynomial : f(x) = x² - ( a + ß)x + aß

Substitute the obtained values in the Formula:

f(x) = x² - ( a + ß)x + aß

→ x² - (8)x + 12

→ x² - 8x + 12

• ∴ f(x) = x² - 8x + 12

Find the zeros of the Quadratic Polynomial:

• f(x) = x² - 8x + 12

→ x² - 8x +12

→ x² - 6x - 2x + 12

→ x(x - 6) - 2(x - 6)

→ (x - 6)(x - 2)

∴ f(x) = (x - 6)(x - 2)

• Therefore, Zeros will be 6 & 2 of the Quadratic Polynomial.

Answer 2

Answer:$\mathfrak{The\ polynomial\ is\ x^2-8x+12=0}$

Step-by-step explanation:

$\rm A\ polynomial\ is\ of\ the\ form\ ax^2-bx+c=0\\ \rm Now\ it's\ given\ that\ the\ sum\ of\ zeroes\ is\ 8\ and\ the\ product\ of\ zeroes\ is\ 12.$

$\rm And\ a\ polynomial\ is\ formed\ by\ x^2-(\alpha+\beta)+(\alpha \times \beta)x=0\\ \rm Let\ the\ two\ zeroes\ be\ \alpha\ and\ \beta\ respectively.$

$We\ know,\ sum\ of\ zeroes= \frac{-b}{a} and\ product\ of\ zeroes= \frac{c}{a}$

$i.e.\ \alpha + \beta = 8\ and\ \alpha \times \beta =12$

$\rm Putting\ values\ in\ the\ equation\\ x^2-(8)x+12=0\\ \rm \implies x^2-8x+12=0\ is\ the\ polynomial.$