
How do you factor and find the zeros of the polynomial $y=-2{{x}^{4}}+10{{x}^{3}}-12{{x}^{2}}$?
Answer
548.1k+ views
Hint: We are given with an expression whose zeroes we have to find. We will start by taking the common terms out. As we can see, we have multiple of 2 in the equation and also we have the multiple of \[{{x}^{2}}\], so we can take these terms out. The remaining equation left inside is solved by using the factoring method. As the degree of the given polynomial is 4, so we will get four values of \[x\].
Complete step by step solution:
According to the given question, we have a polynomial expression which we have to solve for \[x\]. And as we know that degree of polynomial is directly proportional to the number of zeroes of the polynomial.
So for this given expression, we will be having a maximum of 4 zeroes.
We have the expression as,
$y=-2{{x}^{4}}+10{{x}^{3}}-12{{x}^{2}}$-----(1)
We can see that the above expression has a multiple of 2 in each of the terms, we will take 2 common out,
\[\Rightarrow y=-2({{x}^{4}}-5{{x}^{3}}+6{{x}^{2}})\]
Also, we can see in the above expression that it has multiple of \[{{x}^{2}}\], taking that common as well, we get,
\[\Rightarrow y=-2{{x}^{2}}({{x}^{2}}-5x+6)\]-----(2)
Now, we will be solving the quadratic equation \[({{x}^{2}}-5x+6)\], we get,
\[{{x}^{2}}-5x+6\]
\[\Rightarrow {{x}^{2}}-(3+2)x+6\]
\[\Rightarrow {{x}^{2}}-3x-2x+6\]
\[\Rightarrow x(x-3)-2(x-3)\]
Taking the common term out, we get,
\[\Rightarrow (x-3)(x-2)\]----(3)
Putting equation (3) in equation (2), we get,
\[y=-2{{x}^{2}}(x-3)(x-2)\]
We will now equate the equation to 0 to get the zeroes of the given polynomial, we have,
\[-2{{x}^{2}}(x-3)(x-2)=0\]
We have,
\[-2{{x}^{2}}=0\] or \[x-3=0\] and \[x-2=0\]
\[{{x}^{2}}=0\] or \[x=3\] and \[x=2\]
We get the values of \[x=0,3,2\]
Therefore, we get the values of \[x\] are 0,2 and 3.
Note:
We can also solve the above question, by first finding a number using hit and trial method, which when substituted as x gives a value 0.
And we see that when \[x=2\], the expression gives a value 0.
So, we will divide the expression by \[x-2\], we get,
\[x-2\overset{-2{{x}^{3}}+6{{x}^{2}}}{\overline{\left){\begin{align}
& -2{{x}^{4}}+10{{x}^{3}}-12{{x}^{2}} \\
& -(\underline{-2{{x}^{4}}+4{{x}^{3}})} \\
& 0{{x}^{4}}+6{{x}^{3}}-12{{x}^{2}} \\
& -\underline{(0{{x}^{4}}+6{{x}^{3}}-12{{x}^{2}})} \\
& \underline{0{{x}^{4}}+0{{x}^{3}}+0{{x}^{2}}} \\
\end{align}}\right.}}\]
So, we get,
\[y=-2{{x}^{4}}+10{{x}^{3}}-12{{x}^{2}}=(x-2)(-2{{x}^{3}}+6{{x}^{2}})\]
Now, we will equate it to zero, then we have,
\[(x-2)(-2{{x}^{3}}+6{{x}^{2}})=0\]
We have,
\[(x-2)=0\] and \[-2{{x}^{3}}+6{{x}^{2}}=0\]
\[\Rightarrow (x-2)=0,-2{{x}^{3}}+6{{x}^{2}}=0\]
\[\Rightarrow x=2,-2{{x}^{2}}(x-3)=0\]
\[\Rightarrow x=2,-2{{x}^{2}}=0,x-3=0\]
We get the values of \[x=0,x=3,x=2\]
Therefore, the values of \[x\] are 0,2 and 3.
Complete step by step solution:
According to the given question, we have a polynomial expression which we have to solve for \[x\]. And as we know that degree of polynomial is directly proportional to the number of zeroes of the polynomial.
So for this given expression, we will be having a maximum of 4 zeroes.
We have the expression as,
$y=-2{{x}^{4}}+10{{x}^{3}}-12{{x}^{2}}$-----(1)
We can see that the above expression has a multiple of 2 in each of the terms, we will take 2 common out,
\[\Rightarrow y=-2({{x}^{4}}-5{{x}^{3}}+6{{x}^{2}})\]
Also, we can see in the above expression that it has multiple of \[{{x}^{2}}\], taking that common as well, we get,
\[\Rightarrow y=-2{{x}^{2}}({{x}^{2}}-5x+6)\]-----(2)
Now, we will be solving the quadratic equation \[({{x}^{2}}-5x+6)\], we get,
\[{{x}^{2}}-5x+6\]
\[\Rightarrow {{x}^{2}}-(3+2)x+6\]
\[\Rightarrow {{x}^{2}}-3x-2x+6\]
\[\Rightarrow x(x-3)-2(x-3)\]
Taking the common term out, we get,
\[\Rightarrow (x-3)(x-2)\]----(3)
Putting equation (3) in equation (2), we get,
\[y=-2{{x}^{2}}(x-3)(x-2)\]
We will now equate the equation to 0 to get the zeroes of the given polynomial, we have,
\[-2{{x}^{2}}(x-3)(x-2)=0\]
We have,
\[-2{{x}^{2}}=0\] or \[x-3=0\] and \[x-2=0\]
\[{{x}^{2}}=0\] or \[x=3\] and \[x=2\]
We get the values of \[x=0,3,2\]
Therefore, we get the values of \[x\] are 0,2 and 3.
Note:
We can also solve the above question, by first finding a number using hit and trial method, which when substituted as x gives a value 0.
And we see that when \[x=2\], the expression gives a value 0.
So, we will divide the expression by \[x-2\], we get,
\[x-2\overset{-2{{x}^{3}}+6{{x}^{2}}}{\overline{\left){\begin{align}
& -2{{x}^{4}}+10{{x}^{3}}-12{{x}^{2}} \\
& -(\underline{-2{{x}^{4}}+4{{x}^{3}})} \\
& 0{{x}^{4}}+6{{x}^{3}}-12{{x}^{2}} \\
& -\underline{(0{{x}^{4}}+6{{x}^{3}}-12{{x}^{2}})} \\
& \underline{0{{x}^{4}}+0{{x}^{3}}+0{{x}^{2}}} \\
\end{align}}\right.}}\]
So, we get,
\[y=-2{{x}^{4}}+10{{x}^{3}}-12{{x}^{2}}=(x-2)(-2{{x}^{3}}+6{{x}^{2}})\]
Now, we will equate it to zero, then we have,
\[(x-2)(-2{{x}^{3}}+6{{x}^{2}})=0\]
We have,
\[(x-2)=0\] and \[-2{{x}^{3}}+6{{x}^{2}}=0\]
\[\Rightarrow (x-2)=0,-2{{x}^{3}}+6{{x}^{2}}=0\]
\[\Rightarrow x=2,-2{{x}^{2}}(x-3)=0\]
\[\Rightarrow x=2,-2{{x}^{2}}=0,x-3=0\]
We get the values of \[x=0,x=3,x=2\]
Therefore, the values of \[x\] are 0,2 and 3.
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