How do you factor\[250{{x}^{4}}+128x\]?
Answer
580.8k+ views
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.
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.
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