Question

Factorize the given expression.
\[6{{a}^{2}}-a-15\].

Answer 1

Hint: Consider 2 numbers x and y. Take their product as \[\left( -15\times 6 \right)\] and their sum as -1. Thus find the x and y. Split the 2nd term of the expression into 2 terms of x and y and factor out the common terms.

Complete step-by-step answer:

Let us consider two numbers x and y.
Given is the equation \[6{{a}^{2}}-a-15\].
\[6{{a}^{2}}-a-15\] is of the form \[a{{x}^{2}}+bx+c=0\].
Thus the product will be \[\left( a\times c \right)\] and their sum will be taken as b.
Similarly, in \[6{{a}^{2}}-a-15\], the product of the terms x and y will be \[\left( -15\times 6 \right)\] and the sum of the terms will be (-1).
So, the product \[xy=\left( -15\times 6 \right)=-90\].
Similarly the sum, \[x+y=-1\].
Now we need to find the 2 numbers x and y whose products will be (-90) and their sum will be (-1).
If x = -10 and y = 9, let us check if it will satisfy the condition,
\[\begin{align}
  & xy=-10\times 9=-90 \\
 & x+y=-10+9=-1 \\
\end{align}\]
Thus x = -10 and y = 9 satisfy the conditions.
Now let us write down the second term of the expression (-a) as a sum of these two terms whose coefficient are the two numbers,
\[6{{a}^{2}}-a-15=6{{a}^{2}}-10a+9a-15\].
Now let us write the factors out of the first 2 terms and other two terms.
\[6{{a}^{2}}-10a+9a-15=2a(3a-5)+3(3a-5)=(2a+3)(3a-5)\]
Thus we got \[6{{a}^{2}}-10a+9a-15=(2a+3)(3a-5)\].
This can’t be factored further. Thus the factorized value of \[6{{a}^{2}}-10a+9a-15=(2a+3)(3a-5)\].

Note:
This is one of the methods for simple cases. We can also solve this quadratic equation \[6{{a}^{2}}-10a+9a-15\] by comparing with \[a{{x}^{2}}+bx+c=0\].
a = 6, b = -1, c = -15
Substitute these values in \[\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\].
\[a=\dfrac{-(-1)\pm \sqrt{{{(-1)}^{2}}-4\times 6\times (-15)}}{2\times 6}=\dfrac{1\pm \sqrt{1+361}}{12}=\dfrac{1\pm 19}{12}\].
\[\begin{align}
  & a=\dfrac{1+19}{12} \\
 & a=\dfrac{20}{12}=\dfrac{5}{3} \\
 & a-\dfrac{5}{3}=0\Rightarrow 3a-5=0 \\
\end{align}\] and \[\begin{align}
  & a=\dfrac{-18}{12} \\
 & a+\dfrac{3}{2}=0\Rightarrow 2a+3=0 \\
\end{align}\]
Thus the roots of \[6{{a}^{2}}-10a+9a-15=(2a+3)(3a-5)\].