Question

Factorize the following quadratic equation:
\[35{{y}^{2}}+13y-12\]
\[\left( a \right)\left( 7y+3 \right)\left( 5y+4 \right)\]
\[\left( b \right)\left( 7y-3 \right)\left( 5y-4 \right)\]
\[\left( c \right)\left( 7y-3 \right)\left( 5y+4 \right)\]
\[\left( d \right)\left( 7y+3 \right)\left( 5y-4 \right)\]

Answer 1

Hint: We are given a quadratic equation \[35{{y}^{2}}+13y-12.\] We have to find its factor, for that, we will use the middle term split method in which we will split the middle term in such a way that the product of the numbers gives us the number same as the product of the coefficient of \[{{x}^{2}}\left( a \right)\] and the constant term (c). Then, we will take the common and simplify to get the solution.

Complete step-by-step solution:
We are given a quadratic equation \[35{{y}^{2}}+13y-12\] and we are asked to write it as the product of the linear factor. We know that the quadratic equation is defined as \[a{{x}^{2}}+bx+c.\] Comparing it with our equation \[35{{y}^{2}}+13y-12,\] we will get our equation as a quadratic function of x.
Now, to split it into factors we will use the middle term split method. In the middle term split method for quadratic equation, \[a{{x}^{2}}+bx+c,\] we will split the middle term in such a way that the number is given as b (middle term) when added or subtracted while the product of the number gives us the number equal to the product of a and c.
Our equation is \[35{{y}^{2}}+13y-12.\] So, we have a = 35, b = 13 and c = – 12.
We have to split b = 13 in such a way that its product is equal as \[35\times 12=420.\]
We can see that 13 can be written as,
\[13=28-15\]
And,
\[28\times 15=420\]
So, we will use the splitting of 13 as \[13=28-15.\]
Now,
\[35{{y}^{2}}+13y-12=35{{y}^{2}}+\left( 28-15 \right)y-12\]
Simplifying further, we get,
\[\Rightarrow 35{{y}^{2}}+13y-12=35{{y}^{2}}+28y-15y-12\]
Taking the common from the first two terms and the last two terms, we get,
\[\Rightarrow 35{{y}^{2}}+13y-12=7y\left( 5y+4 \right)-3\left( 5y+4 \right)\]
Taking (5y + 4) as common, hence, we get,
\[\Rightarrow 35{{y}^{2}}+13y-12=\left( 7y-3 \right)\left( 5y+4 \right)\]
Therefore, the correct option is (c).

Note: Another way to go for the solution is if we multiply all the options and check which one meets with our question.
We have the equation as \[35{{y}^{2}}+3y-12.\]
\[\left( a \right)\left( 7y+3 \right)\left( 5y+4 \right)\]
Opening the brackets, we get,
\[\left( 7y+3 \right)\left( 5y+4 \right)=7y\left( 5y+4 \right)+3\left( 5y+4 \right)\]
Simplifying further we get,
\[\Rightarrow \left( 7y+3 \right)\left( 5y+4 \right)=35{{y}^{2}}+28y+15y+12\]
\[\Rightarrow \left( 7y+3 \right)\left( 5y+4 \right)=35{{y}^{2}}+43y+12\]
So, the product (7y + 3) (5y + 4) is not equal to our equation.
Hence, (a) is an incorrect option.
\[\left( b \right)\left( 7y-3 \right)\left( 5y-4 \right)\]
Opening the bracket, we get,
\[\left( 7y-3 \right)\left( 5y-4 \right)=7y\left( 5y-4 \right)-3\left( 5y-4 \right)\]
\[\Rightarrow \left( 7y-3 \right)\left( 5y-4 \right)=28{{y}^{2}}-28y-15y+12\]
Simplifying further, we get,
\[\Rightarrow \left( 7y-3 \right)\left( 5y-4 \right)=28{{y}^{2}}-43y+12\]
So, the product (7y – 3) (5y – 4) is not equal to our equation.
Hence, option (b) is an incorrect option.
\[\left( c \right)\left( 7y-3 \right)\left( 5y+4 \right)\]
Opening the bracket, we get,
\[\Rightarrow \left( 7y-3 \right)\left( 5y+4 \right)=7y\left( 5y+4 \right)-3\left( 5y+4 \right)\]
\[\Rightarrow \left( 7y-3 \right)\left( 5y+4 \right)=35{{y}^{2}}+28y-15y-12\]
After simplifying, we get,
\[\Rightarrow \left( 7y-3 \right)\left( 5y+4 \right)=35{{y}^{2}}+13y-12\]
We get, (7y – 3) (5y + 4) is the same as the equation \[\left( 35{{y}^{2}}+13y-12 \right).\]
So, this is the correct option.