Question

How do you factorise \[\left( x+5 \right)\left( x+9 \right)\left( x+3 \right)\left( x+7 \right)-33\]

Answer 1

Hint: Now first we will rearrange the terms if the expression as $\left( x+5 \right)\left( x+7 \right)\left( x+9 \right)\left( x+3 \right)-33$ Now we will use distributive property to solve the first two brackets and the last two brackets. Now we will simplify the expression to form two quadratic expressions. We will further substitute common terms with a variable and solve to form a final quadratic. We will now factor the quadratic in further process.

Complete step by step solution:
Now consider the given expression \[\left( x+5 \right)\left( x+9 \right)\left( x+3 \right)\left( x+7 \right)-33\]
First let us rearrange the terms of the expression. Now we can see that the constants 5 + 7 = 9 + 3. Hence we will use this to simplify the expression. Hence let us first pair such respective terms together.
$\Rightarrow \left( x+5 \right)\left( x+7 \right)\left( x+9 \right)\left( x+3 \right)-33$
First we will try to open the expression and then factor it.
Now we know the distributive property which states $\left( a+c \right)b=ab+cb$ .
Now let us use the property to expand the given expression.
\[\Rightarrow \left[ \left( x+5 \right)\left( x \right)+\left( x+5 \right)\left( 7 \right) \right]\left[ \left( x+9 \right)x+\left( x+9 \right)3 \right]-33\]
Now on simplifying the above expression we get,
\[\begin{align}
  & \Rightarrow \left[ \left( x+5 \right)\left( x \right)+\left( x+5 \right)\left( 7 \right) \right]\left[ \left( x+9 \right)x+\left( x+9 \right)3 \right]-33 \\
 & \Rightarrow \left[ {{x}^{2}}+5x+7x+35 \right]\left[ {{x}^{2}}+9x+3x+27 \right]-33 \\
 & \Rightarrow \left[ {{x}^{2}}+12x+35 \right]\left[ {{x}^{2}}+12x+27 \right]-33 \\
\end{align}\]
Now we can see that both the brackets have ${{x}^{2}}+12x$ common. Hence let us substitute ${{x}^{2}}+12x=a$ .
Hence we get the expression as,
$\begin{align}
  & \Rightarrow \left( a+35 \right)\left( a+27 \right)-33 \\
 & \Rightarrow \left( a+35 \right)a+\left( a+35 \right)27-33 \\
 & \Rightarrow {{a}^{2}}+35a+27a+945-33 \\
 & \Rightarrow {{a}^{2}}+62a-912 \\
\end{align}$
Now we have a quadratic expression in a. Now we will split the middle term of the quadratic in such a way that the product of the terms is the product of the first term and the last term. Hence we will split $62a=24a+38a$ .
Hence we get,
$\begin{align}
  & \Rightarrow {{a}^{2}}+24a+38a-912 \\
 & \Rightarrow a\left( a+24 \right)+38\left( a+24 \right) \\
 & \Rightarrow \left( a+38 \right)\left( a+24 \right) \\
\end{align}$
Now resubstituting the value of a we get,
$\Rightarrow \left[ {{x}^{2}}+12x+24 \right]\left[ {{x}^{2}}+12x+38 \right]$
Now let us solve both the quadratic expression. Now we know that the roots of the quadratic expression are given by $\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ .
Hence using this we get the roots of the expression ${{x}^{2}}+12x+24$ are,
$\begin{align}
  & \Rightarrow x=\dfrac{-12\pm \sqrt{{{12}^{2}}-4\left( 24 \right)}}{2} \\
 & \Rightarrow x=\dfrac{-12\pm \sqrt{144-96}}{2} \\
 & \Rightarrow x=\dfrac{-12\pm 4\sqrt{3}}{2} \\
 & \Rightarrow x=-6\pm 2\sqrt{3} \\
\end{align}$
Hence the factors of the expression are $\left( x-\left( -6-2\sqrt{3} \right) \right)$ and $\left( x-\left( -6+2\sqrt{3} \right) \right)$
Now consider the expression ${{x}^{2}}+12x+38$
The roots of the expression will be,
$\begin{align}
  & \Rightarrow x=\dfrac{-12\pm \sqrt{{{12}^{2}}-4\left( 38 \right)}}{2} \\
 & \Rightarrow x=\dfrac{-12\pm \sqrt{144-152}}{2} \\
 & \Rightarrow x=\dfrac{-12\pm \sqrt{8}i}{2} \\
 & \Rightarrow x=\dfrac{-12\pm 2\sqrt{2}i}{2} \\
 & \Rightarrow x=-6\pm \sqrt{2}i \\
\end{align}$
Hence the factors of the expression are $\left( x-\left( -6+\sqrt{2}i \right) \right)$ and $\left( x-\left( -6-\sqrt{2}i \right) \right)$.

Note: Now note that we can expand the whole expression n and for a polynomial of degree 4 also. Then we will try to guess one root of the polynomial and divide the polynomial with the corresponding factor. Similarly we can do this to form a quadratic polynomial. Finally we factorize the quadratic expression.