Question

How do you factor \[5{{y}^{2}}-12y+4=0\]?

Answer 1

Hint: We are given a quadratic equation which we have to factor. So, the equation we have is in the form, \[a{{y}^{2}}+by+c=0\]. We have the product ‘ac’ as \[20\] and the sum ‘b’ is \[-12\], so we have to factor \[-12\] such that the sum is \[-12\] and the product is \[20\]. We get that as, \[-2\] and \[-10\]. Solving the equation further and taking the common terms out, we will get the factors of the given quadratic equation.

Complete step by step solution:
According to the question given to us, we have a quadratic equation which we have to factor and solve for \[y\].
The quadratic equation we have is,
\[5{{y}^{2}}-12y+4=0\]----(1)
We have the \[Product=ac=5\times 4=20\] and the \[Sum=b=-12\], and so we have to find the terms that will give the product as 20 and the sum as -12.
We have,
\[20=-2\times -10=4\times 5\]
So, the terms we have are -2 and -10. We took -2 and -10 as their product gives a value of 20 and their sum will give the value as -12.
\[\Rightarrow 5{{y}^{2}}+(-10-2)y+4=0\]
Opening up the bracket, we have the expression as,
\[\Rightarrow 5{{y}^{2}}-10y-2y+4=0\]----(3)
Taking the common terms out, we have the expression as,
\[\Rightarrow 5y(y-2)-2(y-2)=0\]
Taking \[(y-2)\] out as common, we get,
\[\Rightarrow (y-2)(5y-2)=0\]
We will now equate these factors to zero and obtain the value of \[y\].
The factors we have are,
\[y-2=0\]
\[\Rightarrow y=2\]
And \[5y-2=0\]
\[\Rightarrow 5y=2\]
\[\Rightarrow y=\dfrac{2}{5}\]

Therefore, the factors of \[(y-2)(y-\dfrac{2}{5})\].

Note: The calculations should be proceeded in a stepwise manner. We can also check whether the values of \[y\] we obtained are correct or not by substituting the values of \[y\] in the given quadratic equation.