Question

Factorise the following expressions.
(i) \[{p^2} + 6p + 8\]
(ii) \[{q^2} - 10q + 21\]
(iii) \[{p^2} + 6p - 16\]

Answer 1

Hint: We will factorize the polynomials using the splitting the middle terms. We will split the middle terms such as the sum of terms is equal to the middle term and the product of terms is equal to the product of the first and last term. Then, take common from the terms and write the polynomial as the product of the factors.

Complete step by step Answer:

We will factorize the given polynomials using splitting the middle term.
In part (i), we have \[{p^2} + 6p + 8\]
We will split the middle term such that the resultant terms add up to $6p$ and their product is $8{p^2}$
Now, we have \[{p^2} + 4p + 2p + 8\]
Take $p$ common from the first two terms and 2 from next two terms.
$p\left( {p + 4} \right) + 2\left( {p + 4} \right)$
We can take $\left( {p + 4} \right)$ common from both the terms.
$\left( {p + 2} \right)\left( {p + 4} \right)$
Therefore, \[{p^2} + 6p + 8\] is factored as $\left( {p + 2} \right)\left( {p + 4} \right)$
Similarly, factorise \[{q^2} - 10q + 21\] for part (ii)
We will split the middle term such that the resultant terms add up to $ - 10q$ and their product is $21{q^2}$
Now, we have \[{q^2} - 7q - 3q + 21\]
Take $q$ common from the first two terms and \[ - 3\] from the next two terms.
$q\left( {q - 7} \right) - 3\left( {q - 7} \right)$
We can take $\left( {q - 7} \right)$ common from both the terms.
$\left( {q - 3} \right)\left( {q + 7} \right)$
Therefore, \[{q^2} - 10q + 21\] is factored as $\left( {q - 3} \right)\left( {q + 7} \right)$
Next, factorise \[{p^2} + 6p - 16\] in part (iii)
We will split the middle term such that the resultant terms add up to $6p$ and their product is $ - 16{p^2}$
Now, we have \[{p^2} - 8q + 2q - 16\]
Take $p$ common from the first two terms and 2 from the next two terms.
$p\left( {p - 8} \right) + 2\left( {p - 8} \right)$
We can take $\left( {p - 8} \right)$ common from both the terms.
$\left( {p + 2} \right)\left( {p - 8} \right)$
Therefore, \[{p^2} + 6p - 16\] is factored as $\left( {p + 2} \right)\left( {p - 8} \right)$

Note: If we multiply the factors of the polynomial, then we will get back the original polynomial. One should take care of the sign of the middle term while splitting it. We can also use the factorization to calculate the zeroes of the polynomial by putting each of the factors equal to 0.