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

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.