Question

How do you factor\[250{{x}^{4}}+128x\]?

Answer 1

Hint: In this type of question, at the time of solving we should know a formula that is \[{{a}^{3}}+{{b}^{3}}=\left( a+b \right)\left( {{a}^{2}}+{{b}^{2}}-ab \right)\]. Also, we should know the cubes of some specific numbers like\[{{1}^{3}}=1\], \[{{4}^{2}}=64\], \[{{5}^{3}}=125\], etc. In this question, first we will factor out the x. After that, we factor the remaining equation using the formula that is \[{{a}^{3}}+{{b}^{3}}=\left( a+b \right)\left( {{a}^{2}}+{{b}^{2}}-ab \right)\].

Complete step by step answer:
Let us solve this question.
It is asked in the question that how do we factor the equation\[250{{x}^{4}}+128x\].
So, we have to factorize the equation\[250{{x}^{4}}+128x\].
So, for the factorization of the above equation, we will first factor out the x because it can be seen from the equation that x is a common factor in that equation.
Therefore, \[250{{x}^{4}}+128x=250\times x\times x\times x\times x+128\times x\]
 x is common here in the above. So, that factor can be taken out.
\[\Rightarrow 250{{x}^{4}}+128x=\left( 250\times x\times x\times x+128 \right)\times x\]
\[\Rightarrow 250{{x}^{4}}+128x=\left( 250\times {{x}^{3}}+128 \right)\times x\]
\[\Rightarrow 250{{x}^{4}}+128x=x\left( 250{{x}^{3}}+128 \right)\]
Similarly, 2 can be a factor out from the above equation.
\[\Rightarrow 250{{x}^{4}}+128x=x\left( 2\times 125\times {{x}^{3}}+2\times 64 \right)\]
\[\Rightarrow 250{{x}^{4}}+128x=2x\left( 125{{x}^{3}}+64 \right)\]
Now, from seeing the above equation, we can write the values
\[125{{x}^{3}}=5\times 5\times 5\times x\times x\times x={{5}^{3}}\times {{x}^{3}}={{\left( 5x \right)}^{3}}\]
And
\[64=4\times 4\times 4={{4}^{3}}\]
\[\Rightarrow 250{{x}^{4}}+128x=2x\left( {{\left( 5x \right)}^{3}}+{{4}^{3}} \right)\]
Here, we have got addition of two perfect cubes.
As we know that the addition of two perfect cubes that is \[{{a}^{3}}+{{b}^{3}}\] can be factorised into two factors that will be \[\left( {{a}^{2}}+{{b}^{2}}-ab \right)\] and \[\left( a+b \right)\]
Hence, \[{{a}^{3}}+{{b}^{3}}\] can be written as the multiplication of \[\left( {{a}^{2}}+{{b}^{2}}-ab \right)\] and \[\left( a+b \right)\].
Therefore, \[{{a}^{3}}+{{b}^{3}}=\left( a+b \right)\left( {{a}^{2}}+{{b}^{2}}-ab \right)\]
So, we can write
\[\left( 125{{x}^{3}}+64 \right)=\left( {{\left( 5x \right)}^{3}}+{{4}^{3}} \right)=\left( 5x+4 \right)\left( {{\left( 5x \right)}^{2}}+{{4}^{2}}-5x\times 4 \right)\]
\[\Rightarrow \left( 125{{x}^{3}}+64 \right)=\left( 5x+4 \right)\left( 25{{x}^{2}}+16-20x \right)\]
The factor of \[\left( 25{{x}^{2}}+16-20x \right)\] cannot be done because it has no roots and no factors also.
Therefore, the factors of \[\left( 125{{x}^{3}}+64 \right)\] are \[\left( 5x+4 \right)\] and \[\left( 25{{x}^{2}}+16-20x \right)\]
Hence, the factorization of \[250{{x}^{4}}+128x\] will be \[2x\left( 5x+4 \right)\left( 25{{x}^{2}}+16-20x \right)\].

Note:
 For solving this type of question, we should have a better knowledge of polynomials. So, we can solve these types of questions easily. Don’t forget to check whether \[\left( 25{{x}^{2}}+16-20x \right)\] has factors or not. Otherwise, the solution may be wrong.
Method to check the factors of \[\left( 25{{x}^{2}}+16-20x \right)\] :
As the above equation is a quadratic polynomial. If it can be factorized, then it will have roots.
\[25{{x}^{2}}+16-20x=25{{x}^{2}}-20x+16\]
Hence, for the roots to be real, its discriminant value should be positive.
As we know that the discriminant of any general quadratic equation \[a{{x}^{2}}+bx+c=0\] is \[D=\sqrt{{{b}^{2}}-4ac}\]
The discriminant, D= \[{{(-20)}^{2}}-4\times 25\times 16\]
\[{{(-20)}^{2}}-4\times 25\times 16=400-1600=-1200\]
As it is negative, hence the equation has no roots. Therefore, it cannot be factorized.
So, we can use the above method to check if the quadratic equation has the factors or not.