The zeroes of quadratic polynomial are x²+25x+156 are

1.	The zeroes of quadratic polynomial are x²+25x+156 are
Answer» Answer :- Method I :- By middle term splitting method :- → x² + 25X + 156 → x² + 12x + 13x + 156 → x ( x + 12 ) + 13 ( x + 12 ) → ( x + 12 ) ( x + 13 ) → x + 12 = 0 x = -12 → x + 13 = 0 x = -13 Method II :- Using quadratic formula :- $\implies\sf x = \dfrac{-b\pm \sqrt{b^2 - 4ac}}{2a}$ For the GIVEN equation :- a = 1 b = 25 c = 156 $\implies\sf x = \dfrac{-25 \pm \sqrt{25^2 - 4 \times 156 \times 1}}{2 \times 1}$ $\implies\sf x = \dfrac{-25 \pm \sqrt{625 - 624}}{2}$ $\implies\sf x = \dfrac{-25 \pm \sqrt1}{2}$ $\implies\sf x = \dfrac{-25 \pm 1}{2}$ $\implies\sf x = \dfrac{-25 + 1}{2} , \dfrac{-25 - 1}{2}$ $\implies\sf x = \dfrac{-24}{2} , \dfrac{-26}{2}$ $\implies\sf x = - 12 , -13$ Zeros of the polynomial = -12 , -13

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