factorize : x²-5x+6
Answers
Step-by-step explanation:
Thanks to ask this question.
It pretty simple when you get to understand the method to find the factors of such polynomials.
The methods by which you can solve such polynomials for their factors are:
Factorization Method
By Shri Dharacharya’s Quadratic Formula.
Method of perfect Square
*METHOD-1*
Factorization method:
Given: x^2+5x+6=0
So as you know this is a quadratic equation, and it would posses two roots. Hence you need two find two numbers in this method who has a sum of 5 and a product of 6.
Looking to the requirement, we can have a following no. of possibilties for two number to have a sum of 5 and a product 6. Which could be
0 and 5, here the sum is 5 but the product is 0, hence this is not the one we are looking for.
1 and 4, here the sum is 5, but the product is 4, hence it is also not the required case.
2 and 3, here the sum is 5, and the product is 6, bravo these are the two numbers which we were looking for.
Hence now rewrite the above equation in the terms of the two numbers 2 and 3.
x^2 + (2+3)x +(2×3)=0
x^2 + 2x + 3x +(2×3)=0
Now here you can see the first two terms have x common in them and the last two terms have 3 common in them, take them out as below.
x(x + 2) + 3(x + 2)=0
Now you can see (x + 2) is common in the two terms of the equation, take it out as below.
(x + 2)(x + 3)=0
Here we come with two possiblities, as (x + 2)=0 and (x + 3)=0.
From here we have the two roots of the the above equation, which are x = -2 and x = -3.
So the answer is x = -2, -3 are the required roots.
*METHOD-2*
Shri Dharacharya’s Quadratic Formula:
To know this formula, we must first know the standard form of a quadratic equation, which is ax^2 + bx + c=0, where a, b, c are the coefficients.
And the Formula is framed as per the standard form, which is as below.
x = [{-b±√(b^2 -4ac)}/2a]
So, by using this formula you can easily find the solution.
Now comparing your given equation, we have a=1, b=5 and c=6.
Putting these values in the formula, x= [{-5±√(5^2–4×1×6)}/(2×1)]
x=[{-5±√(25–24)}/2]
x=[{-5±√1}/2]
x=[{-5±1}/2]Now again we have two possibilities.
x= [{-5+1}/2] or x=[{-5–1}/2]
x=[-4/2] or x=[{-6/2}]
x=-2 or x=-3
Bravo, we reached to the solution, hence roots of given equation are x= -2,-3
*METHOD-3*
Method of perfect square:
In this particular method, we try to make a perfect square, and then find solution by taking root of it, you will get to know this when I will solve this problem by this method.
Given: x^2 + 5x + 6=0
Rewriting x^2 + {2×(5/2)×x} +6=0
Now adding and subtracting square of (5/2) . x^2 + {2×(5/2)×x} + {(5/2)^2} -{(5/2)^2} +6=0
Now you can see the first three term are like the expression of the square of some expression, for example a^2 + 2ab + b^2 which is a square of (a + b).
So here we can write our expression as {x + (5/2)}^2 - (25/4)+6=0
Further solving {x + (5/2)}^2 + {(-25+24)/4}=0
{x + (5/2)}^2 + (-1/4)=0
{x + (5/2)}^2 -(1/4)=0
Taking 1/4 on the other side. {x + (5/2)}^2 = (1/4)
Taking square root on both sides {x + (5/2)} = ±(1/2)
Now we have two possibilities, {x + (5/2)}=(1/2) and {x + (5/2)}= (-1/2)
On evaluating we have, x= (1/2)-(5/2) and x=(-1/2)+(-5/2)
x= -2 and x= -3
So we have the solution to the equation as x= -2, -3 from this method also.
So you saw that these three method gave the same solution. And I hope you have understood how two find the solution to a quadratic equation, by these three methods, now you can easily find the solution of any quadratic equation by any of the three methods.
So, plz mark me as brainliest answer .