Chapter 11, Problem 10GR

To determine

Solve the equation $12y^2-y=35$ by factoring.

Answer to Problem 10GR

Solutions of the equation $12y^2-y=35$ are $y=-\frac{5}{3}\text{and}y=\frac{7}{4}$

Explanation of Solution

Given:

The equation: $12y^2-y=35$

Concept Used:

Zero-Product Property: If we multiply two (or more) things together and the result is equal to zero, then we know that  at least one  of those things that we multiplied must also have been equal to zero.

Calculation:

Rewrite the equation in standard form: $12y^2-y=35\Rightarrow 12y^2-y-35=0$

Middle Term Factorization:

In Quadratic Factorization using Splitting of Middle Term which is x term is the sum of two factors and product equal to last term.

To Factor the form : $a{x}^2 +bx+c$	Factor : $12y^2-y-35$
Find the product of 1st and last terms ( $a·c$).	$12·-35=-420$
2) Find the factors of $-420$ in such way that addition or subtraction of that factors is the middle term ( $-y$ )(Splitting of middle term)	$-420=-21·20$ and $-21y+20y=-y$
3) Write the middle term using the sum of the two new factors, including the proper signs.	$\begin{array}{l}12y^2-y-35\\ 12y^2-21y+20y-35\end{array}$
4) Group the terms to form pairs - the first two terms and the last two terms. Factor each pair by finding common factors.	$\begin{array}{l}12y^2-21y+20y-35\\ 3y(4y-7)+5(4y-7)\end{array}$
5) Factor out the shared (common) binomial parenthesis.	( $12y^2-y-35=(3y+5)(4y-7)$

Now use zero product property:

  $\begin{array}{l}12y^2-y-35=0\\ (3y+5)(4y-7)0\\ 3y+5=0\text{and}4y-7=0[\text{Zero-Product Property}]\\ y=-\frac{5}{3}\text{and}y=\frac{7}{4}\end{array}$

Thus, Solutions of the equation $12y^2-y=35$ are $y=-\frac{5}{3}\text{and}y=\frac{7}{4}$