
Factorize \[9{x^2} - 25\]
Answer
563.7k+ views
Hint: We have given a quadratic polynomial and we have to find its factor. Firstly we put it equal to \[p\left( x \right)\]. Then we write the factor of each term of polynomial. Then we write them in square form. This will be equivalent to the identity \[{a^2} - {b^2}\]. The identity \[{a^2} - {b^2}\] is equal to \[\left( {a + b} \right)\left( {a - b} \right)\]. So we apply this identity on the polynomial.
Since the polynomial is quadratic so it will have two factors.
Complete answer:
We have given a quadratic polynomial. Let it be equal to \[p\left( x \right)\].
$\Rightarrow$\[p\left( x \right) = 9{x^2} - 25\]
We have to find the factor of the polynomial. Now \[9{x^2}\]can be written as \[3 \times 3 \times x \times x\] and \[3 \times 3 \times x \times x\] can be written as \[{\left( {3x} \right)^2}\]. Also \[25\] can be written as \[5 \times 5\] and \[5 \times 5\] and \[5 \times 5\] can be written as \[{\left( 5 \right)^2}\]
$\Rightarrow$ \[p\left( x \right) = {\left( {3x} \right)^2} - {\left( 5 \right)^2}\]
$\Rightarrow$\[{\left( {3x} \right)^2} - {\left( 5 \right)^2}\] is in the form \[{a^2} - {b^2}\] and \[{a^2} - {b^2}\] is equal to \[\left( {a + b} \right)\left( {a - b} \right)\]
$\Rightarrow$ \[{\left( {3x} \right)^2} - {\left( 5 \right)^2}\] will be equal to \[\left( {3x + 5} \right)\left( {3x - 5} \right)\]
$\Rightarrow$ \[p\left( x \right)\]\[ = \left( {3x + 5} \right)\left( {3x - 5} \right)\]
This is the factorized form of \[9{x^2} - 25\].
Note: Polynomials which have degree two are called quadratic polynomials. The quadratic polynomial is also called a second order polynomial. It is in the form \[p\left( x \right) = a{x^2} + bx + c\] where a should not be equal to zero. A quadratic polynomial has degree two so it has two factors and two roots. The quadratic polynomial given here is univariate because it has only one variable. There are also bivariate polynomials which have degree 2 and two variables. The bivariate polynomial is written as
\[p\left( {x,y} \right) = a{x^2} + b{y^2} + cxy + dx + ey\] with at least one \[a,b,c\] not equal to equal to zero.
Since the polynomial is quadratic so it will have two factors.
Complete answer:
We have given a quadratic polynomial. Let it be equal to \[p\left( x \right)\].
$\Rightarrow$\[p\left( x \right) = 9{x^2} - 25\]
We have to find the factor of the polynomial. Now \[9{x^2}\]can be written as \[3 \times 3 \times x \times x\] and \[3 \times 3 \times x \times x\] can be written as \[{\left( {3x} \right)^2}\]. Also \[25\] can be written as \[5 \times 5\] and \[5 \times 5\] and \[5 \times 5\] can be written as \[{\left( 5 \right)^2}\]
$\Rightarrow$ \[p\left( x \right) = {\left( {3x} \right)^2} - {\left( 5 \right)^2}\]
$\Rightarrow$\[{\left( {3x} \right)^2} - {\left( 5 \right)^2}\] is in the form \[{a^2} - {b^2}\] and \[{a^2} - {b^2}\] is equal to \[\left( {a + b} \right)\left( {a - b} \right)\]
$\Rightarrow$ \[{\left( {3x} \right)^2} - {\left( 5 \right)^2}\] will be equal to \[\left( {3x + 5} \right)\left( {3x - 5} \right)\]
$\Rightarrow$ \[p\left( x \right)\]\[ = \left( {3x + 5} \right)\left( {3x - 5} \right)\]
This is the factorized form of \[9{x^2} - 25\].
Note: Polynomials which have degree two are called quadratic polynomials. The quadratic polynomial is also called a second order polynomial. It is in the form \[p\left( x \right) = a{x^2} + bx + c\] where a should not be equal to zero. A quadratic polynomial has degree two so it has two factors and two roots. The quadratic polynomial given here is univariate because it has only one variable. There are also bivariate polynomials which have degree 2 and two variables. The bivariate polynomial is written as
\[p\left( {x,y} \right) = a{x^2} + b{y^2} + cxy + dx + ey\] with at least one \[a,b,c\] not equal to equal to zero.
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