Question

factorize by splitting the middle term
$y{}^{2} + 2y - 5 \div 4$
​

Answer 1

Given equation

$y^{2} +2y-\dfrac{5}{4} =0$

Multiplying the equation by 4, we get,

$4y^{2} +8y-5=0$

Solving, we get,

$4y^{2} -8y-5=0$

$\implies 4y^{2} -10y+2y-5=0$

$\implies 2y(2y-5)+1(2y-5)=0$

$\implies (2y+1)(2y-5)=0$

Now,

$2y+1=0\\\\\ \implies y = \dfrac{-1}{2}$

$2y-5=0\\\ \implies y = \dfrac{5}{2}$

$\boxed{\boxed{\bold{Therefore, \ the \ factorised \ expression \ is \ (2y+1)(2y-5)}}}}}$