How do you factor ${{f}^{3}}-4{{f}^{2}}-9f+36$?
Answer
608.1k+ views
Hint: Now to factorize the given expression we will first take ${{f}^{2}}$ common from first two terms and – 9 common from last two terms. Now we will simplify the equation. Now again we will simplify the equation by using the formula ${{a}^{2}}-{{b}^{2}}=\left( a-b \right)\left( a+b \right)$ hence using this and simplifying the obtained equation we get all the factors of the given expression.
Complete step by step solution:
Now we are given a polynomial in f of degree 3.
Now since there is no direct method to factorize a degree 3 polynomial we will have to simplify the given expression.
Now to factorize the polynomial we will first group common terms.
Now consider the given polynomial ${{f}^{3}}-4{{f}^{2}}-9f+36$
Now to factorize the polynomial let us take ${{f}^{2}}$ common from the first two terms and $-9$ common from the last two terms.
$\Rightarrow {{f}^{2}}\left( f-4 \right)-9\left( f-4 \right)$
Now taking $\left( f-4 \right)$ common from the above expression we get,
$\Rightarrow \left( {{f}^{2}}-9 \right)\left( f-4 \right)$
Now we have a quadratic equation ${{f}^{2}}-9$
To factorize the expression we will use the formula ${{a}^{2}}-{{b}^{2}}=\left( a-b \right)\left( a+b \right)$ here we have a = f and b = 3. Hence we get $\left( {{f}^{2}}-9 \right)=\left( f+3 \right)\left( f-3 \right)$ .
Now substituting this in the above equation we get the given equation as,
$\Rightarrow \left( f-3 \right)\left( f+3 \right)\left( f-4 \right)$
Hence the factors of the given equation are $\left( f+3 \right)$ , $\left( f-3 \right)$ and $\left( f-4 \right)$ .
Note: Now note that if we have $\left( x-\alpha \right)$ as a factor of the equation in x then $x=\alpha $ will be the root of the equation. Hence we can easily find one root of the equation and hence find one factor. Now we can divide the equation by the factor to simplify. Now we can factorize the quadratic equation. Also note that we can also find the roots of the obtained quadratic and then find factors of the expression.
Complete step by step solution:
Now we are given a polynomial in f of degree 3.
Now since there is no direct method to factorize a degree 3 polynomial we will have to simplify the given expression.
Now to factorize the polynomial we will first group common terms.
Now consider the given polynomial ${{f}^{3}}-4{{f}^{2}}-9f+36$
Now to factorize the polynomial let us take ${{f}^{2}}$ common from the first two terms and $-9$ common from the last two terms.
$\Rightarrow {{f}^{2}}\left( f-4 \right)-9\left( f-4 \right)$
Now taking $\left( f-4 \right)$ common from the above expression we get,
$\Rightarrow \left( {{f}^{2}}-9 \right)\left( f-4 \right)$
Now we have a quadratic equation ${{f}^{2}}-9$
To factorize the expression we will use the formula ${{a}^{2}}-{{b}^{2}}=\left( a-b \right)\left( a+b \right)$ here we have a = f and b = 3. Hence we get $\left( {{f}^{2}}-9 \right)=\left( f+3 \right)\left( f-3 \right)$ .
Now substituting this in the above equation we get the given equation as,
$\Rightarrow \left( f-3 \right)\left( f+3 \right)\left( f-4 \right)$
Hence the factors of the given equation are $\left( f+3 \right)$ , $\left( f-3 \right)$ and $\left( f-4 \right)$ .
Note: Now note that if we have $\left( x-\alpha \right)$ as a factor of the equation in x then $x=\alpha $ will be the root of the equation. Hence we can easily find one root of the equation and hence find one factor. Now we can divide the equation by the factor to simplify. Now we can factorize the quadratic equation. Also note that we can also find the roots of the obtained quadratic and then find factors of the expression.
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