Question

find the zeroes of factorising polynomial p(x) = 4x2 - 12x +5 and verify the relationship between the zeroes and the coefficient​

Answer 1

Solution :

We have quadratic polynomial p(x) = 4x² - 12x + 5

Zero of the polynomial p(x) = 0

$\longrightarrow\sf{4x^{2} -12x+5=0}\\\\\longrightarrow\sf{4x^{2} -2x-10x+5=0}\\\\\longrightarrow\sf{2x(2x-1)-5(2x-1)=0}\\\\\longrightarrow\sf{(2x-1)(2x-5)=0}\\\\\longrightarrow\sf{2x-1=0\:\:\:Or\:\:\:2x-5=0}\\\\\longrightarrow\sf{2x=1\:\:\:Or\:\:\:2x=5}\\\\\longrightarrow\bf{x=1/2\:\:\:Or\:\:\:x=5/2}$

∴ α = 1/2 & β = 5/2 are the zeroes of the polynomial.

As we know that given polynomial compared with ax² + bx + c;

• a = 4
• b = -12
• c = 5

Now;

$\underline{\mathcal{SUM\:OF\:THE\:ZEROES\::}}$

$\mapsto\sf{\alpha +\beta =\dfrac{-b}{a} =\bigg(\dfrac{Coefficient\:of\:x }{Coefficient\:of\:x^{2} }\bigg)}\\\\\\\mapsto\sf{\dfrac{1}{2} +\dfrac{5}{2} =\dfrac{-(-12)}{4} }\\\\\\\mapsto\sf{\dfrac{1+5}{2} =\cancel{\dfrac{12}{4}}}\\\\\\\mapsto\bf{\dfrac{6}{2} =\dfrac{6}{2} }$

$\underline{\mathcal{PRODUCT\:OF\:THE\:ZEROES\::}}$

$\mapsto\sf{\alpha \times \beta =\dfrac{c}{a} =\bigg(\dfrac{Constant\:term }{Coefficient\:of\:x^{2} }\bigg)}\\\\\\\mapsto\sf{\dfrac{1}{2} \times \dfrac{5}{2} =\dfrac{5}{4} }\\\\\\\mapsto\bf{\dfrac{5}{4} =\dfrac{5}{4} }$

Thus;

The zeroes & coefficient of relationship are verified .