Question

How do you factor \[2{x^2} + 8x + 4\]?

Answer 1

Hint: A polynomial of degree two is called a quadratic polynomial and its zeros can be found using many methods like factorization, completing the square, graphs, quadratic formula etc. The quadratic formula is used when we fail to find the factors of the equation. If factors are difficult to find then we use Sridhar’s formula to find the roots. That is \[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\].

Complete step-by-step solution:
Given, \[2{x^2} + 8x + 4\]
Since the degree of the equation is 2, we have 2 factors.
On comparing the given equation with the standard quadratic equation\[a{x^2} + bx + c\], we have\[a = 2\], \[b = 8\] and \[c = 4\].
The standard form of the factorization of quadratic equation is \[a{x^2} + {b_1}x + {b_2}x + c\], which satisfies the condition \[{b_1} \times {b_2} = a \times c\] and \[{b_1} + {b_2} = b\].
This is the case with factorization method files.
So we use quadratic formula,
\[ \Rightarrow x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\], substituting we have,
\[ \Rightarrow x = \dfrac{{ - 8 \pm \sqrt {{8^2} - 4\left( 2 \right)\left( 4 \right)} }}{{2\left( 2 \right)}}\]
\[ \Rightarrow x = \dfrac{{ - 8 \pm \sqrt {64 - 32} }}{4}\]
\[ \Rightarrow x = \dfrac{{ - 8 \pm \sqrt {32} }}{4}\]
We can write 32 as a product of 16 and 2
\[ \Rightarrow x = \dfrac{{ - 8 \pm \sqrt {16 \times 2} }}{4}\]
\[ \Rightarrow x = \dfrac{{ - 8 \pm 4\sqrt 2 }}{4}\]
Taking 4 common we have,
\[ \Rightarrow x = \dfrac{{4\left( { - 2 \pm \sqrt 2 } \right)}}{4}\]
\[ \Rightarrow x = - 2 \pm \sqrt 2 \]
Thus the roots are
\[ \Rightarrow x = - 2 + \sqrt 2 \] and \[x = - 2 - \sqrt 2 \].

Hence the factors are \[ \Rightarrow x - \left( { - 2 + \sqrt 2 } \right)\] and \[x - \left( { - 2 - \sqrt 2 } \right)\].

Note: We can check whether the obtained solution is correct or wrong. All we need to do is substituting the value of ‘y’ in the given problem.
\[\Rightarrow y - 8 = 11\]
\[\Rightarrow 19 - 8 = 11\]
\[ \Rightarrow 11 = 11\]
Hence the obtained answer is correct.
In the above we did the transpose of addition and subtraction. Similarly if we have multiplication we use division to transpose. If we have division we use multiplication to transpose. Follow the same procedure for these kinds of problems.