Question

Factorise \[54{x^2} + 42{x^3} - 30{x^4}\]

Answer 1

Hint: Factoring is the process in which an expression or number is expressed as the product of its factors, that is by factorising we obtain numbers or terms which can be multiplied to get back the original expression. We will factorise the equation by taking a common term.

Complete step-by-step answer:
The given expression is a polynomials which are algebraic expressions that are composed of two or more algebraic terms, the algebraic terms are constant, variables and exponents, there are different types of polynomials which are differentiated with the help of degree of the polynomial.
The polynomial of degree 4 is known as biquadratic polynomial. The given polynomial is a biquadratic polynomial.
Now given equation is \[54{x^2} + 42{x^3} - 30{x^4}\] ,
The coefficients of the given polynomial are \[54,42,30\]
As we know using simple mathematics the common factor all these numbers are \[6\]
And we also see the terms of polynomial without coefficients are \[{x^2},{x^3},{x^4}\]
Similarly using simple mathematics the common factor of these will be \[{x^2}\]
Therefore the common factor of the whole polynomial will be \[6 \times {x^2} = 6{x^2}\]
Now let us take this term common from the polynomial and write it as follows
\[ \Rightarrow 54{x^2} + 42{x^3} - 30{x^4} = 6{x^2}\left( { - 5{x^2} + 7x + 9} \right)\]
But we can still factorise the second term \[ - 5{x^2} + 7x + 9\]
It is in the form of a quadratic equation \[a{x^2} + bx + c\] where \[a = - 5\], \[b = 7\] and \[c = 9\],
Taking values of a,b and c in \[{b^2} - 4ac\] \[ = {\left( 7 \right)^2} - \left( {4 \times - 5 \times 9} \right)\] \[ = 49 + 180 = 229 > 0\] this means the equation has real roots .
No using formula for finding roots \[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\]
\[ = \dfrac{{ - 7 \pm \sqrt {229} }}{{ - 10}}\]
\[ = \dfrac{{7 + \sqrt {229} }}{{10}}\] or \[\dfrac{{7 - \sqrt {229} }}{{10}}\]
Therefore the polynomial gets simplified as \[\therefore 6{x^2}\left( {x - \left( {\dfrac{{7 + \sqrt {229} }}{{10}}} \right)} \right)\left( {x - \left( {\dfrac{{7 - \sqrt {229} }}{{10}}} \right)} \right)\]
\[\therefore 54{x^2} + 42{x^3} - 30{x^4} = 6{x^2}\left( {x - \left( {\dfrac{{7 + \sqrt {229} }}{{10}}} \right)} \right)\left( {x - \left( {\dfrac{{7 - \sqrt {229} }}{{10}}} \right)} \right)\]
So, the correct answer is “\[6{x^2}\left( {x - \left( {\dfrac{{7 + \sqrt {229} }}{{10}}} \right)} \right)\left( {x - \left( {\dfrac{{7 - \sqrt {229} }}{{10}}} \right)} \right)\]”.

Note: Factorization is a process which is necessary to simplify the algebraic expressions and is used to solve the higher degree equations. It is the inverse procedure of the multiplication of the polynomials. The algebraic expression is said to be in a factored form only when the whole expression is an indicated product.