Question

find the zeros of the polynomial 4x²+7x-2​

Answer 1

GIVEN :-

• p(x) = 4x² + 7x - 2.

TO FIND :-

• The zeros of p(x).

SOLUTION :-

☛ We have to find the zeros of the polynomial so, p(x) = 0.

☯ ACCORDING TO QUESTION,

➫ p(x) = 0

➫ 4x² + 7x - 2 = 0

➫ 4x² + 7x - 2 = 0

➫ 4x² + 7x -2 = 0

➫ 4x² - x + 8x - 2 = 0

➫ x(4x - 1) + 2(4x - 1) = 0

➫ (4x - 1) , (x + 2) = 0

➫ 4x - 1 = 0 , x + 2 = 0

➫ 4x = 1 , x = -2

➫ x = 1/4 , x = -2

Hence the zeros of p(x) are 1/4 and (-2).

Answer 2

ANSWER✔

$\large\underline\bold{GIVEN,}$

$\sf\dashrightarrow 4x^2+7x-2$

$\large\underline\bold{TO\:FIND,}$

$\sf\dashrightarrow \:ZEROES\:OF\:THE\:POLYNOMIAL.$

ACCORDING TO THE QUESTION,

$\sf {\boxed{\bf{\:\: to\:find\:zeroes\:of\:the\:polynomial\:we\:need\:p(x)=0}}}$

$\large\underline\bold{SOLTUION,}$

$\sf\implies 4x^2+7x-2=0$

$\sf\implies 4x^2+8x-x-2=0$

$\sf\implies 4x(x+2)-1(x+2)=0$

$\sf\implies (4x-1)(x+2)=0$

$\sf\implies 4x-1=0 ,x+2=0$

$\sf\implies 4x=1,x=-2$

$\sf\implies x= \dfrac{1}{4},x=-2$

$\sf\implies x= \dfrac{1}{4},-2$

$\large{\boxed{\bf{ \star\:\:x= \dfrac{1}{4},-2\:\: \star }}}$

$\large\underline\bold{THE\:ZEROES\:OF\:THE\:POLYNOMIAL\:ARE\: \dfrac{1}{4},-2}$

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