Question

How do you factor the equation $y=-2{{x}^{2}}+3x+5$ into y = (x – p) (x – q) ?

Answer 1

Hint: We have factor $y=-2{{x}^{2}}+3x+5$ which is a quadratic equation. To factor a quadratic equation $a{{x}^{2}}+bx+c$ we split the box into 2 terms such that product of coefficients of each term is equal to ac. In this case ac is equal to -10. We can split 3x into - 2x + 5x . Then we can take -2x common from $-2{{x}^{2}}-2x$ and 5 common form 5x + 5 .

Complete step by step solution:
The given equation is $y=-2{{x}^{2}}+3x+5$
If we compare it with $a{{x}^{2}}+bx+c$ we get a = - 2, b = 3, and c = 5
The value of ac is - 10
So, we have to find 2 numbers such that their sum is 3 and product is – 10. The numbers are 5 and – 2
So, we can write
 $\Rightarrow -2{{x}^{2}}+3x+5=-2{{x}^{2}}-2x+5x+5$
We can take – 2x common from first 2 terms and 5 common from last 2 terms
$\Rightarrow -2{{x}^{2}}+3x+5=-2x\left( x+1 \right)+5\left( x+1 \right)$
Now we can take x + 1 common from the whole equation
$\Rightarrow -2{{x}^{2}}+3x+5=\left( -2x+5 \right)\left( x+1 \right)$
We can take – 2 common from -2x + 5
 $\Rightarrow -2{{x}^{2}}+3x+5=-2\left( x-\dfrac{5}{2} \right)\left( x+1 \right)$
$-2\left( x-\dfrac{5}{2} \right)\left( x+1 \right)$ is the factored form of the equation $y=-2{{x}^{2}}+3x+5$

Note: In quadratic equation $a{{x}^{2}}+bx+c$ , ${{b}^{2}}-4ac$ is discriminant of the equation. If the discriminant is less than 0, then we can not factor an equation because in the factor form a (x – f) (x – g) . f and g are roots of the equation. if discriminant is less than 0, there will be no real roots, so factorization is not possible.