Question

Find the zeroes of the polynomial 21x2 – 11x –2 = 0

Answer 1

Answer :-

Given :-

• Polynomial :- 21x² - 11x - 2 = 0

To Find :-

• Zeros / roots

Solution :-

Method 1 :-

Using the middle term splitting method :-

$\implies\sf 21x^2 - 11x - 2 = 0$

$\implies\sf 21x^2 - 14x + 3x - 2 = 0$

$\implies\sf 7x ( 3x - 2 ) + 1 ( 3x - 2 ) = 0$

$\implies\sf ( 7x + 1 ) ( 3x - 2 ) = 0$

Either ( 7x + 1 ) = 0 or ( 3x - 2 ) = 0

$\implies\sf 7x + 1 = 0$

$\implies\sf 7x = -1$

$\implies\sf x = \dfrac{1}{7}$

$\implies\sf 3x - 2 = 0$

$\implies\sf 3x = 2$

$\implies\sf x = \dfrac{2}{3}$

Zeros of the polynomial = ⅔ , - ⅐

Method 2 :-

Using quadratic formula :-

$\implies\sf Zeros = \dfrac{-b \pm \sqrt{b^2-4ac}}{2a}$

• a = 21
• b = -11
• c = -2

$\implies\sf \dfrac{-(-11) \pm \sqrt{11^2 - 4 \times 21 \times (-2)}}{2\times 21}$

$\implies\sf \dfrac{11 \pm \sqrt{121 + 168}}{2\times 21}$

$\implies\sf \dfrac{11 \pm \sqrt{289}}{42}$

$\implies\sf \dfrac{11 \pm 17}{42}$

$\implies\sf \dfrac{11+17}{42} , \dfrac{11-17}{42}$

$\implies\sf \dfrac{28}{42} , \dfrac{-6}{42}$

$\implies\sf \dfrac{2}{3} , \dfrac{-1}{7}$

Zeros of the polynomial = ⅔ , - ⅐

Answer 2

Hello Mate

Here's your answer

Hope you find it easy