Question

Factorise $ \left( {x + 1} \right)\left( {x + 2} \right)\left( {x + 3} \right)\left( {x + 4} \right) - 120 $ .

Answer 1

Hint: In the given problem, we have to factorise the given polynomial and represent it as a product of its factors. The given polynomial is of degree $ 4 $ and is thus called a bi-quadratic polynomial. Bi-quadratic polynomials are also called quartic polynomials. For factoring the polynomial, first we have to simplify the given polynomial and then find its factors.

Complete step-by-step answer:
So, the given polynomial is $ \left( {x + 1} \right)\left( {x + 2} \right)\left( {x + 3} \right)\left( {x + 4} \right) - 120 $ . Simplifying and rearranging,
 $ \left( {x + 1} \right)\left( {x + 2} \right)\left( {x + 3} \right)\left( {x + 4} \right) - 120 $ $ = $ $ \left[ {x\left( {x + 2} \right) + 1(x + 2)} \right]\left( {x + 3} \right)\left( {x + 4} \right) - 120 $
 $ = $ $ \left[ {\left( {{x^2} + 2x + x + 2} \right)} \right]\left( {x + 3)(x + 4} \right) - 120 $ $ $
 $ = $ $ \left[ {\left( {{x^2} + 3x + 2} \right)} \right]\left( {x + 3} \right)(x + 4) - 120 $
On further simplifying,
 $ = $ \[\left[ {x\left( {{x^2} + 3x + 2} \right) + 3\left( {{x^2} + 3x + 2} \right)} \right]\left( {x + 4} \right) - 120\]
 $ = $ \[\left[ {{x^3} + 3{x^2} + 2x + 3{x^2} + 9x + 6} \right]\left( {x + 4} \right) - 120\]
Simplifying the polynomial,
 $ = $ \[\left[ {{x^3} + 6{x^2} + 11x + 6} \right]\left( {x + 4} \right) - 120\]
Opening the bracket,
 $ = $ $ x\left[ {{x^3} + 6{x^2} + 11x + 6} \right] + 4\left[ {{x^3} + 6{x^2} + 11x + 6} \right] - 120 $
 $ = $ $ \left[ {{x^4} + 6{x^3} + 11{x^2} + 6x} \right] + \left[ {4{x^3} + 24{x^2} + 44x + 24} \right] - 120 $
Adding the brackets,
 $ = $ $ \left[ {{x^4} + 10{x^3} + 35{x^2} + 50x + 24} \right] - 120 $
 $ = $ $ {x^4} + 10{x^3} + 35{x^2} + 50x - 96 $
So, $ p\left( x \right) $ $ = $ $ {x^4} + 10{x^3} + 35{x^2} + 50x - 96 $
Now, using hit and trial method, consider given polynomial as $ p\left( x \right) $
Putting $ x = 1 $ in the polynomial $ p\left( x \right) $ ,
 $ = $ \[p(x = 1) = {\left( 1 \right)^4} + 10{\left( 1 \right)^3} + 35{\left( 1 \right)^2} + 50\left( 1 \right) - 96\]
 $ = $ $ p\left( {x = 1} \right) = 1 + 10 + 35 + 50 - 96 $
On simplifying further,
 $ = $ $ p\left( {x = 1} \right) = 96 - 96 $
 $ = $ $ p\left( {x = 1} \right) = 0 $
So, $ p(1) = 0 $ .
So, using factor theorem, $ \left( {x - 1} \right) $ is a factor of p(x).
Now, $ p(x) = (x - 1)({x^3} + 11{x^2} + 46x + 96) $
Further applying hit and trial method for factorising $ ({x^3} + 11{x^2} + 46x + 96) $ .
Let $ f(x) = ({x^3} + 11{x^2} + 46x + 96) $ .
Putting $ x = - 6 $ in the polynomial, we observe that,
 $ f(x = - 6) = {( - 6)^3} + 11{( - 6)^2} + 46( - 6) + 96 $
 $ = $ $ f(x = - 6) = - 216 + 396 - 276 + 96 $
 $ = $ $ f(x = - 6) = 0 $
By hit and trial, we observe that $ - 6 $ is a zero of $ ({x^3} + 11{x^2} + 46x + 96) $ .
We observe that $ - 6 $ is a zero of $ ({x^3} + 11{x^2} + 46x + 96) $ .
So, using the factor theorem, $ \left( {x + 6} \right) $ is a factor of $ ({x^3} + 11{x^2} + 46x + 96) $ .
Therefore,
 $ p(x) = \left( {x - 1} \right)\left( {x + 6} \right)\left( {{x^2} + 5x + 16} \right) $ $ $
Now quadratic polynomial $ \left( {{x^2} + 5x + 16} \right) $ cannot be factorised further as its Discriminant is less than $ 0 $ . The quadratic equations having discriminant less than $ 0 $ don’t have any real roots.
Therefore,
 $ p(x) = \left( {x - 1} \right)\left( {x + 6} \right)\left( {{x^2} + 5x + 16} \right) $
So, the correct answer is “ $ p(x) = \left( {x - 1} \right)\left( {x + 6} \right)\left( {{x^2} + 5x + 16} \right) $ ”.

Note: Such polynomials can be factorised by using the hit and trial method and then using factor theorem finding the factors of the polynomials. Bi-quadratic polynomials are polynomials with degree $ 4 $ . We can convert such bi-quadratic polynomials into quadratic polynomials by finding out $ 2 $ factors of the polynomial by hit and trial and then easily factorising the quadratic polynomial with splitting of the middle term or using Quadratic formula.