Question

Explain the Gemetrical meaning of zero of polynomial Also find the zero of t square - 15 .Also verify relationship between zeros and coefficients?​

Answer 1

Answer:

zero polynomial equal degree of polynomial is zero example 100,40,35,36

these are example because there was no variable there was only constant drum so the degree of polynomial is zero and known as zero polynomial

Answer 2

FINAL ANSWER:-

t=$\sqrt{15}$ and t=$-\sqrt{15}$

GIVEN:-

polynomial is $t^{2}-15$

TO FIND:-

• the Geometrical meaning of zero of polynomial

• verify relationship between zeros and coefficients

• the zero of $t^{2}-15$

THINGS TO KNOW:-

polynomial structure ⇒ $ax^{2}+bx+c$

sum of zeroes = $\frac{-b}{a}$ → $\alpha +\beta =\frac{-b\:\:(coefficient \:of\:x )}{a\:\:(coefficient\:of\:x^{2})}$

product of zeroes = $\frac{c}{a}$ → $\alpha \beta =\frac{c\:\:(constant)}{a\:\:(coefficient\:of\:x^{2} )}$

SOLUTION:-

• The zero of the polynomial is the x-coordinate of the point, where the graph intersects the x-axis. If a polynomial p(x) intersects the x-axis at ( k, 0), then k is the zero of the polynomial.

• t² - 15 = 0

       ⇒t² - (√15)² = 0

      ⇒(t + √15)(t - √15) = 0

      ⇒t = √15 , - √15

      zeroes of polynomial t² - 15 are :-  α=$\sqrt{15}$    ;     β=$-\sqrt{15}$

• relationship between zeros and coefficients :-

here polynomial structure ⇒$ax^{2}+bx+c$   t² - 15

a=1

b=0

c=-15

∴sum of zeroes(α , β) ⇒ $\alpha +\beta =\frac{-b}{a}$

$\sqrt{15}-\sqrt{15}=\frac{0}{1}$ ⇒ 0=0

∴product of zeroes ⇒ $\alpha \beta =\frac{c}{a}$

$\sqrt{15}*-\sqrt{15}=\frac{-15 }{1}$ ⇒ $-15=-15$