Question

Find the zeroes of the following polynomials by factorization method and verify
the relations between the zeroes and coefficients of the polynomials. 4x^2- 3x - 1
​

Answer 1

Answer:

here is ur answer....

hope it helps...

Answer 2

Solution :

We have quadratic polynomial p(x) = 4x² - 3x - 1

Zero of the polynomial p(x) = 0

A/q

$\longrightarrow\sf{4x^{2}-3x-1=0}\\\\\longrightarrow\sf{4x^{2} -4x+x -1=0}\\\\\longrightarrow\sf{4x(x-1)+1(x-1)=0}\\\\\longrightarrow\sf{(x-1)(4x+1)=0}\\\\\longrightarrow\sf{x-1=0\:\:\:Or\:\:\:4x+1=0}\\\\\longrightarrow\sf{x=1\:\:\:Or\:\:\:4x=-1}\\\\\longrightarrow\bf{x=1\:\:\:Or\:\:\:x=-1/4}$

∴α = 1 & β = -1/4 are the zeroes of the polynomial .

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

• a = 4
• b = -3
• c = -1

Now;

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

$\mapsto\sf{\alpha +\beta =\dfrac{-b}{a} =\bigg\lgroup \dfrac{Coefficient\:of\:x}{Coefficient\:of\:x^{2} } \bigg\rgroup}\\\\\\\mapsto\sf{1+\bigg(-\dfrac{1}{4} \bigg)=\dfrac{-(-3)}{4} }\\\\\\\mapsto\sf{1-\dfrac{1}{4} =\dfrac{3}{4} }\\\\\\\mapsto\sf{\dfrac{4-1}{4} =\dfrac{3}{4} }\\\\\\\mapsto\bf{\dfrac{3}{4} =\dfrac{3}{4}}$

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

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

Thus;

The zeroes & relations between zeroes & coefficient are verified .