
How do you simplify (x + 10) (3x – 5)?
Answer
497.1k+ views
Hint: We will first use the fact that (a + b)(c + d) = a(c + d) + b(c + d) and then we will use the distributive property to obtain the required answer.
Complete step-by-step answer:
We are given that we are required to simplify (x + 10) (3x – 5).
We know that (a + b)(c + d) = a(c + d) + b(c + d) for any real numbers a, b, c and d.
On replacing a by x, b by 10, c by 3x and d by – 5, we will then obtain:-
$ \Rightarrow $(x + 10) (3x – 5) = x(3x – 5) + 10(3x – 5) ……………….(1)
Now, we will use the distributive property for three real number x, y and z which states that:
x(y + z) = xy + xz ………………(2)
Replacing x by nothing, y by 3x and z by – 5 in the equation number 2, we will then obtain the following expression:-
$ \Rightarrow $ x(3x – 5) = x(3x) – 5x
Simplifying the expression in the above line, we will then obtain the following expression:-
$ \Rightarrow x(3x - 5) = 3{x^2} - 5x$ ……………(3)
Replacing x by 10, y by 3x and z by – 5 in the equation number 2, we will then obtain the following expression:-
$ \Rightarrow $ 10 (3x – 5) = 10 (3x) – 5(10)
Simplifying the expression in the above line, we will then obtain the following expression:-
$ \Rightarrow $ 10 (3x – 5) = 30x - 50 ……………(4)
Putting the equation number (3) and (4) in equation number (1), we will then obtain the following expression:-
$ \Rightarrow (x + 10)(3x - 5) = 3{x^2} - 5x + 30x - 50$
Now, we will club both the terms with x in the right hand side of the above expression to obtain the following expression:-
$ \Rightarrow (x + 10)(3x - 5) = 3{x^2} + 25x - 50$
Note:
The students must note that we mentioned the term real numbers in the property mentioned above but this holds true for complex numbers as well, there we mentioned the real numbers because we used the real numbers in the solution only.
Complete step-by-step answer:
We are given that we are required to simplify (x + 10) (3x – 5).
We know that (a + b)(c + d) = a(c + d) + b(c + d) for any real numbers a, b, c and d.
On replacing a by x, b by 10, c by 3x and d by – 5, we will then obtain:-
$ \Rightarrow $(x + 10) (3x – 5) = x(3x – 5) + 10(3x – 5) ……………….(1)
Now, we will use the distributive property for three real number x, y and z which states that:
x(y + z) = xy + xz ………………(2)
Replacing x by nothing, y by 3x and z by – 5 in the equation number 2, we will then obtain the following expression:-
$ \Rightarrow $ x(3x – 5) = x(3x) – 5x
Simplifying the expression in the above line, we will then obtain the following expression:-
$ \Rightarrow x(3x - 5) = 3{x^2} - 5x$ ……………(3)
Replacing x by 10, y by 3x and z by – 5 in the equation number 2, we will then obtain the following expression:-
$ \Rightarrow $ 10 (3x – 5) = 10 (3x) – 5(10)
Simplifying the expression in the above line, we will then obtain the following expression:-
$ \Rightarrow $ 10 (3x – 5) = 30x - 50 ……………(4)
Putting the equation number (3) and (4) in equation number (1), we will then obtain the following expression:-
$ \Rightarrow (x + 10)(3x - 5) = 3{x^2} - 5x + 30x - 50$
Now, we will club both the terms with x in the right hand side of the above expression to obtain the following expression:-
$ \Rightarrow (x + 10)(3x - 5) = 3{x^2} + 25x - 50$
Note:
The students must note that we mentioned the term real numbers in the property mentioned above but this holds true for complex numbers as well, there we mentioned the real numbers because we used the real numbers in the solution only.
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