Question

How do you factor the expression \[4{{z}^{2}}+32z+63\]?

Answer 1

Hint: In this problem, we have to find the factor of the given expression by factorization method. We can first split the middle term, i.e. the x term with its coefficient. We have to expand the middle term in such a way that their addition is equal to the middle term i.e. 32z, and multiplication is equal to \[4\times 63=252=18\times 14\], where \[18+14=32\] which is the middle term. We can then take common terms outside to get the factors.

Complete step by step solution:
We know that the given expression is,
\[4{{z}^{2}}+32z+63\]
We can first split the middle term to form factors.
We have to expand the middle term in such a way that their addition is equal to the middle term i.e. 32z, and multiplication is equal to \[4\times 63=252=18\times 14\], where \[18+14=32\] which is the middle term.
\[\Rightarrow 4{{z}^{2}}+18z+14z+63\]
We can now take the first two terms and the last two terms to take common terms outside, we get
\[\Rightarrow \left( 4{{z}^{2}}+18z \right)+\left( 14z+63 \right)\]
Now we can take the common terms outside, we get
\[\Rightarrow 2z\left( 2z+9 \right)+7\left( 2z+9 \right)\]
We can again take the common factor first then the remaining terms, to make a factor, we get
\[\Rightarrow \left( 2z+9 \right)\left( 2z+7 \right)\]

Therefore, the factors are \[\left( 2z+9 \right)\left( 2z+7 \right)\].

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,
\[4{{z}^{2}}+32z+63\]
Where, a = 4, b = 32, c = 63.
We can substitute the above values in quadratic formula.
\[\begin{align}
  & \Rightarrow z=\dfrac{-32\pm \sqrt{1024-1008}}{8} \\
 & \Rightarrow z=\dfrac{-32\pm \sqrt{16}}{8} \\
 & \Rightarrow z=\dfrac{-8\pm 4}{18} \\
 & \Rightarrow x=-\dfrac{7}{2},-\dfrac{9}{2} \\
\end{align}\]
Where,\[z=-\dfrac{9}{2},x=-\dfrac{7}{2}\]
We can take the term in the right-hand side, to the left-hand side by changing its sign.
\[\Rightarrow \left( 2z+9 \right)\left( 2z+7 \right)\]
Therefore, the factors are \[\left( 2z+9 \right)\left( 2z+7 \right)\].