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Question: How do you factor the trinomial \[4{x^2} + 20x + 25 = 0\]?...

How do you factor the trinomial 4x2+20x+25=04{x^2} + 20x + 25 = 0?

Explanation

Solution

Factoring reduces the higher degree equation into its linear equation. In the above question, we need to reduce the trinomial into its simplest form in such a way that addition of products of the factors of first and last term should be equal to the middle term.

Complete step by step solution:
Trinomial is a polynomial consisting of three terms or monomials. The above equation is a trinomial since it has three terms.
ax2+bx+ca{x^2} + bx + c is a general way of writing quadratic equations where a, b and c are the numbers.
In the above expression,
a=4, b=20, c=25
4x2+20x+25=04{x^2} + 20x + 25 = 0
First step is by multiplying the coefficient of 4x24{x^2} and the constant term 25, we get 100x2100{x^2}.
After this, factors of 100x2100{x^2} should be calculated in such a way that their addition should be equal to 20x20x.
Factors of 100 can be 10 and 10.
where 10x+10x=20x10x + 10x = 20x.
So, further we write the equation by equating it with zero and splitting the middle term according to the factors.

4x2+20x+25=0 4x2+10x+10x+25=0 \Rightarrow 4{x^2} + 20x + 25 = 0 \\\ \Rightarrow 4{x^2} + 10x + 10x + 25 = 0 \\\

Now, by grouping the first two and last two terms we get common factors.

2x(2x+5)+5(2x+5)=0 (2x+5)(2x+5)=0 \Rightarrow 2x\left( {2x + 5} \right) + 5\left( {2x + 5} \right) = 0 \\\ \Rightarrow \left( {2x + 5} \right)(2x + 5) = 0 \\\

Taking 2x common from the first group and 5 common from the second we get the above equation.
We can further solve it and taking square root we get,

(2x+5)2=0  {\left( {2x + 5} \right)^2} = 0 \\\ \\\

Therefore, we get the above solution for the equation.

Note: In quadratic equation, an alternative way of finding the factors is by using the formula of sum of the terms a and b (a+b)2=a2+2ab+b2{\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2}.So by
converting the equation according to the formula we get (2x)2+2×2x×5+(5)2=(2x+5)2{(2x)^2} + 2 \times 2x \times 5 + {(5)^2} = {(2x + 5)^2}