Question

How do you factor \[5{{y}^{2}}-5y-30\]?

Answer 1

Hint: This type of problem is based on the concept of factoring a polynomial. First, we have to consider the polynomial with degree 2. First, we have to take 5 common from all the terms of the polynomial. We have to split the middle term of the polynomial in such a way that we get common terms from the first and last term. Here, the middle term is -1y. We can split the middle as an addition of 3y and -2y. Then, we need to take y common from the first two terms and 2 common from the last two terms. We find that (y-3) is common in the two terms. On taking the common terms, we get the factors of the polynomial.

Complete step by step solution:
According to the question, we are asked to find the factors of \[5{{y}^{2}}-5y-30\].
We have been given the polynomial is \[5{{y}^{2}}-5y-30\]. ---------(1)
The given polynomial is of degree 2 and variable y.
Here, we find that 5 are common in all the terms of the considered polynomial. On taking 5
common, we get
\[5{{y}^{2}}-5y-30=5\left( {{y}^{2}}-y-6 \right)\]
Now, let us factorise \[{{y}^{2}}-y-6\].
To find the factors, we have to consider the middle term and split the middle term in such a
way that the addition of these two terms will be -1 and multiplication will be -6.
We know that -3+2=-1 and \[-3\times 2=-6\].
Therefore, we get
\[5{{y}^{2}}-5y-30=5\left( {{y}^{2}}+\left( -3+2 \right)y-6 \right)\]
Using the distributive property in the middle term, that is \[a\left( b+c \right)=ab+ac\].
Here, a=y, b=-3 and c=2.
\[\Rightarrow 5{{y}^{2}}-5y-30=5\left( {{y}^{2}}+\left( -3 \right)y+2y-6 \right)\]
\[\Rightarrow 5{{y}^{2}}-5y-30=5\left( {{y}^{2}}-3y+2y-6 \right)\]
We can express the polynomial as
\[5{{y}^{2}}-5y-30=5\left( {{y}^{2}}-3y+2y-3\times 2 \right)\]
We find that y is common in the first two terms of the simplified polynomial and 2 are
common in the last two terms of the simplified polynomial.
Let us take y and 2 common out of the bracket respectively.
\[\Rightarrow 5{{y}^{2}}-5y-30=5\left( y\left( y-3 \right)+2\left( y-3 \right) \right)\]
We find that y-3 is common in both the terms of the equation. On taking (y-3) common, we
get
\[5{{y}^{2}}-5y-30=5\left( \left( y-3 \right)\left( y+2 \right) \right)\]
\[\Rightarrow 5{{y}^{2}}-5y-30=5\left( y-3 \right)\left( y+2 \right)\]
Here, we find that the polynomial is converted as a product of two linear polynomials which
are the factors of the given polynomial.
Therefore, the factors of \[5{{y}^{2}}-5y-30\] are 5(y-3) and y+2.

Note: Whenever we get such a type of problem, we have to split the middle terms to find the factors. We can also write the factors as y-3 and 5(y+2). 5 can be multiplied to any factor and the final answer can be obtained. Avoid calculation mistakes based on sign convention.