Question

How do you factor the expression \[18{{x}^{2}}-15x-18=0\]?

Answer 1

Hint: In this problem, we have to find the factor of the given expression by factorization method. We can first take the common term outside to factorize. We can then factorize the equation by expanding the middle term i.e. the n term with its coefficient in such a way that their addition is equal to the middle term and multiplication is equal to the multiplication of first term and the last term. We can then take common terms outside to get the factors.

Complete step by step solution:
We know that the given expression is,
\[18{{x}^{2}}-15x-18=0\]
We can now take the common term outside, we get
\[\Rightarrow 3\left( 6{{x}^{2}}-5x-6 \right)\]
We can now factorize the above equation.
We can first split the middle term to form factors.
We have to expand the middle term i.e. the x term with its coefficient in such a way that their addition is equal to the middle term i.e. -5x, and multiplication is equal to \[-6\times 6=-36=-9\times 4\]
\[\begin{align}
  & \Rightarrow -6\times 6=-36=-9\times 4 \\
 & \Rightarrow -9+4=-5 \\
\end{align}\]
We can apply the above step,
\[\Rightarrow 3\left( 6{{x}^{2}}+4x-9x-6 \right)\]
We can now take the first two terms and the last two terms to take common terms outside, we get
\[\Rightarrow 3\left[ \left( 6{{x}^{2}}+4x \right)-\left( 9x+6 \right) \right]\]
Now we can take the common terms outside, we get
\[\Rightarrow 3\left[ 2x\left( 3x+2 \right)-3\left( 3x+2 \right) \right]\]
We can again take the common factor first then the remaining terms to make a factor, we get
\[\Rightarrow 3\left( 2x-3 \right)\left( 3x+2 \right)\]

Therefore, the factors are \[3\left( 2x-3 \right)\left( 3x+2 \right)\].

Note: We can also verify for the correct answer using the quadratic formula.
The quadratic formula for the equation \[a{{x}^{2}}+bx+c\] is
\[x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\]
We know that the given expression is,
\[18{{x}^{2}}-15x-18=0\]
We can now take the common term outside, we get
\[\Rightarrow 3\left( 6{{x}^{2}}-5x-6 \right)\]
Where, a = 6, b = -5, c = -6.
We can substitute the above values in quadratic formula.
\[\begin{align}
  & \Rightarrow x=\dfrac{5\pm \sqrt{25+144}}{12} \\
 & \Rightarrow x=\dfrac{5\pm 13}{12} \\
 & \Rightarrow x=\dfrac{3}{2},-\dfrac{2}{3} \\
\end{align}\]
Where,\[x=-\dfrac{2}{3},x=\dfrac{3}{2}\]
We can take the term in the right-hand side, to the left-hand side by changing its sign.
\[\Rightarrow 3\left( 2x-3 \right)\left( 3x+2 \right)\]
Therefore, the factors are \[3\left( 2x-3 \right)\left( 3x+2 \right)\].