How do you write \[{x^2} - 6x + 9\] in factored form?
Answer
625.5k+ views
Hint: Here, we will find the factors of the quadratic equation by using the concept of factorization. First, we will split the middle term of the equation, and then we will form the factors by taking the common terms in the equation to get the equation in the factored form.
Complete step by step solution:
We are given a Quadratic equation \[{x^2} - 6x + 9\].
The standard form of a quadratic equation is \[a{x^2} + bx + c\].
Now, comparing the given quadratic equation with the general quadratic equation, so we get
\[\begin{array}{l}a = 1\\b = - 6\\c = 9\end{array}\]
Here, we will split the middle term in such a way that the number must be equal to the product of the coefficient of the first term and the constant.
Now the product of the first term and last term is equal to
\[ac = \left( 1 \right)\left( 9 \right) = 9\]
Now we can see that \[ - 3\] and \[ - 3\] when multiplied will be equal to the product of the first term and last term and when they are added they are equal to the middle term.
Thus, we will split the middle term in the given quadratic equation as \[ - 3x\] and \[ - 3x\] . So, we get
\[{x^2} - 6x + 9 = {x^2} - 3x - 3x + 9\]
By taking out the common factors, we get
\[ \Rightarrow {x^2} - 6x + 9 = x\left( {x - 3} \right) - 3\left( {x - 3} \right)\]
By grouping the common factors, we get
\[ \Rightarrow {x^2} - 6x + 9 = \left( {x - 3} \right)\left( {x - 3} \right)\].
Therefore, \[\left( {x - 3} \right)\] and \[\left( {x - 3} \right)\] are the factors of \[{x^2} - 6x + 9\].
Note:
We know that Factorization is a process of rewriting the expression in terms of the product of the factors. Factors are numbers if the expression is a numeral. Factors are algebraic expressions if the expression is an algebraic expression. Factorization is done by using the common factors and the grouping of terms. We should remember that if the sign of the product of the coefficient of the first term and constant is Positive, then both the integers should be either positive or both must be negative but if the sign of the sum of the factors is Negative, then both the integers should be negative.
Complete step by step solution:
We are given a Quadratic equation \[{x^2} - 6x + 9\].
The standard form of a quadratic equation is \[a{x^2} + bx + c\].
Now, comparing the given quadratic equation with the general quadratic equation, so we get
\[\begin{array}{l}a = 1\\b = - 6\\c = 9\end{array}\]
Here, we will split the middle term in such a way that the number must be equal to the product of the coefficient of the first term and the constant.
Now the product of the first term and last term is equal to
\[ac = \left( 1 \right)\left( 9 \right) = 9\]
Now we can see that \[ - 3\] and \[ - 3\] when multiplied will be equal to the product of the first term and last term and when they are added they are equal to the middle term.
Thus, we will split the middle term in the given quadratic equation as \[ - 3x\] and \[ - 3x\] . So, we get
\[{x^2} - 6x + 9 = {x^2} - 3x - 3x + 9\]
By taking out the common factors, we get
\[ \Rightarrow {x^2} - 6x + 9 = x\left( {x - 3} \right) - 3\left( {x - 3} \right)\]
By grouping the common factors, we get
\[ \Rightarrow {x^2} - 6x + 9 = \left( {x - 3} \right)\left( {x - 3} \right)\].
Therefore, \[\left( {x - 3} \right)\] and \[\left( {x - 3} \right)\] are the factors of \[{x^2} - 6x + 9\].
Note:
We know that Factorization is a process of rewriting the expression in terms of the product of the factors. Factors are numbers if the expression is a numeral. Factors are algebraic expressions if the expression is an algebraic expression. Factorization is done by using the common factors and the grouping of terms. We should remember that if the sign of the product of the coefficient of the first term and constant is Positive, then both the integers should be either positive or both must be negative but if the sign of the sum of the factors is Negative, then both the integers should be negative.
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