Question

How do you factor $9{x^2} - 42x + 49$

Answer 1

Hint: This problem comes under factoring polynomial expression into factors on algebra. The general form of quadratic expression is $a{x^2} + bx + c$. There are methods to find factors of polynomial factors; they are factorisation method, and division root method. Here we use the factorisation method. Here there will be polynomial expression of algebraic identity on comparing the general form of expansion of quadratic expression with their co-efficient and substitute the values in the formula we will get required factors. By using formula and complete step by step explanation we solve this.

Formula used: ${(a - b)^2} = {a^2} - 2ab + {b^2}$

Complete step-by-step solution:
Let us consider the quadratic equation $9{x^2} - 42x + 49$
Now on factorising comparing coefficient of ${x^2}$, $x$ and constant term in general form of quadratic equation $a{x^2} + bx + c = 0$ with $9{x^2} - 42x + 49$ , we get
\[a = 9,b = - 42,c = 49\]
Now sum of roots is\[b = - 42\]
The product of roots is $a \times c = 49 \times 9$
The product of root is \[26x + 97 = 15 \times 13\]
Therefore we need to find the factors where the sum of two numbers is $ - 42$ and the product of two numbers is $441$
The numbers are $ - 21, - 21$
The sum of numbers is $ - 21 - 21 = - 42$
The product of numbers is $ - 21 \times - 21 = 441$
Now we need to factorisation we have separate the sum of roots: $9{x^2} - 21x - 21x + 49$
Now take common numbers and variables on the expression: $3x(3x - 7) - 7(3x - 7)$
$ \Rightarrow (3x - 7)(3x - 7)$

Therefore the factors are $(3x - 7)(3x - 7)$

Note: The polynomial factorisation is to express the given polynomial into the factors. Here the quadratic polynomial has to be factored into two factors. Here we solve by factorisation method. We can solve by the division method GCD (Greatest Common Divisor) method and also if the given quadratic polynomial is perfect square we can solve using the square root method. Here we solve by factorisation method because the sum and product of roots are little simple to find the factors.