Question

How do you factor $ {{x}^{4}}+10{{x}^{2}}+9 $ ?

Answer 1

Hint: In this question, we are given an expression of degree 4 and we need to factorize it. For factorizing it easily, we will put $ {{x}^{2}} $ as y and form a quadratic equation in terms of y. After that, we will apply a split middle term method to factorize the expression. At last, we will substitute $ {{x}^{2}} $ for y and then factorize the obtained factors. Further, we will use the property that $ {{i}^{2}}=-1 $ .

Complete step by step answer:
Here we are given the expression as $ {{x}^{4}}+10{{x}^{2}}+9 $ .
We need to factorize it. For simplification let us substitute $ {{x}^{2}} $ as y, so $ {{x}^{2}}=y $ .
Squaring both sides we get $ {{x}^{4}}={{y}^{2}} $ .
So our expression becomes $ {{y}^{2}}+10y+9 $ .
As we can see, the new expression is in the quadratic form so let us use the split the middle term method to factorize it.
We know, for an equation of the form $ a{{x}^{2}}+bx+c $ to apply split the middle term method we need to find two terms $ {{n}_{1}}\text{ and }{{n}_{2}} $ such that their product is equal to the product of a and c and their sum is equal to b i.e $ {{n}_{1}}+{{n}_{2}}=b\text{ and }{{n}_{1}}\cdot {{n}_{2}}=a\cdot c $ .
Comparing $ {{y}^{2}}+10y+9 $ with $ a{{x}^{2}}+bx+c $ we have a = 1, b = 10 and c = 9.
So we need $ {{n}_{1}}\text{ and }{{n}_{2}} $ such that $ {{n}_{1}}+{{n}_{2}}=10\text{ and }{{n}_{1}}\cdot {{n}_{2}}=9 $ .
Factors of 9 are 1, 3, 9. We can see, only 1 and 9 are possible numbers such that 1+9 = 10 and $ 1\times 9=9 $ . Therefore, we will write 10y as y+9y. We get $ {{y}^{2}}+y+9y+9 $ .
Taking y common from the first two terms and a from the last two terms we get $ y\left( y+1 \right)+9\left( y+1 \right) $ .
Now let us take (y+1) common we get $ \left( y+1 \right)\left( y+9 \right) $ .
Substituting $ {{x}^{2}} $ for y we get $ \left( {{x}^{2}}+1 \right)\left( {{x}^{2}}+9 \right) $ .
Therefore, we have obtained two factors. For factorising further we can use $ \left( {{x}^{2}}-\left( -1 \right) \right)\left( {{x}^{2}}-\left( -9 \right) \right) $ .
We know that, $ {{\left( 3i \right)}^{2}}=9{{i}^{2}}=-9\text{ and }{{\left( i \right)}^{2}}=-1 $ .
So we can write $ \left( {{x}^{2}}-{{i}^{2}} \right)\left( {{x}^{2}}-{{\left( 3i \right)}^{2}} \right) $ .
Applying $ {{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right) $ we get $ \left( x+i \right)\left( x-i \right)\left( x+3i \right)\left( x-3i \right) $ .
Therefore we have obtained four factors of $ {{x}^{4}}+10{{x}^{2}}+9 $ .

Note:
Students should take care of the signs while factorizing using i. They can leave their answer for two factors only if four factors are not asked but try to factorize as much as possible. Select $ {{n}_{1}}\text{ and }{{n}_{2}} $ carefully which satisfies both requirements.