How do you factor ${x^2} - 3x - 10$ ?
Answer
573.3k+ views
Hint: We will try to factor the given equation by splitting the middle term. Find the two factors of the constant term in the equation whose sum is equal to the coefficient of the middle term. Then, take the common factors and make two pairs.
Complete step by step solution:
We have been given the equation ${x^2} - 3x - 10$.
Therefore, from the above equation, we can see that –
The first term is ${x^2}$, and its coefficient is 1.
The second term of the equation is $ - 3x$ whose coefficient is $ - 3$.
The last term is the constant which is -10.
We have to split the middle term so, that we can share common factors and make two pairs and then make each of them equal to zero. As the factors of any number are those which give the remainder as 0.
So, to apply the method of splitting the middle term the steps which we have to apply are –
Multiply the coefficient of the first term, ${x^2}$ which is 1 with the last term or the constant term which is -10: $1. - 10 = - 10$.
Now, the product of the coefficient of the first term and last term is -10. Therefore, find the two factors of -10 whose sum equals the coefficient of the middle term, which is -3.
If we multiply -10 and 1, we’ll get the product as -10 but the sum will be -9.
If we multiply -5 and 2, we’ll get the product as -10 and the sum is -3.
Hence, we’ll use -5 and 2 in splitting the middle term.
As we know, the sum of -5 and 2 is equal to 3. Hence, rewriting the polynomial, we get –
$ \Rightarrow {x^2} - 5x + 2x - 10$
Now, pulling out like factors from the first two terms in the above equation, we get –
$ \Rightarrow x\left( {x - 5} \right) + 2x - 10$
Taking common factor in the last terms in the above equation, we get –
$ \Rightarrow x\left( {x - 5} \right) + 2\left( {x - 5} \right)$
Hence, the equation becomes - $\left( {x - 5} \right)\left( {x + 2} \right)$
We know that factors of any number are those which when divided that number gives the remainder as 0.
Therefore, in the equation –
$ \Rightarrow \left( {x - 5} \right)\left( {x + 2} \right) = 0$
$
\Rightarrow \left( {x - 5} \right) = 0 \\
\therefore x = 5 \\
$
And –
$
\Rightarrow x + 2 = 0 \\
\therefore x = - 2 \\
$
Hence, this is the required factorization.
Note:
We can also solve this question, by using the Quadratic formula which is –
$x = \dfrac{{ - B \pm \sqrt {{B^2} - 4AC} }}{{2A}}$
Here, $A$ is the coefficient of the first term, $B$ is the coefficient of the second term or middle term and $C$ is the constant term.
Putting the values of $A, B$, and $C$ in the Quadratic formula will give us the exact factorization.
Complete step by step solution:
We have been given the equation ${x^2} - 3x - 10$.
Therefore, from the above equation, we can see that –
The first term is ${x^2}$, and its coefficient is 1.
The second term of the equation is $ - 3x$ whose coefficient is $ - 3$.
The last term is the constant which is -10.
We have to split the middle term so, that we can share common factors and make two pairs and then make each of them equal to zero. As the factors of any number are those which give the remainder as 0.
So, to apply the method of splitting the middle term the steps which we have to apply are –
Multiply the coefficient of the first term, ${x^2}$ which is 1 with the last term or the constant term which is -10: $1. - 10 = - 10$.
Now, the product of the coefficient of the first term and last term is -10. Therefore, find the two factors of -10 whose sum equals the coefficient of the middle term, which is -3.
If we multiply -10 and 1, we’ll get the product as -10 but the sum will be -9.
If we multiply -5 and 2, we’ll get the product as -10 and the sum is -3.
Hence, we’ll use -5 and 2 in splitting the middle term.
As we know, the sum of -5 and 2 is equal to 3. Hence, rewriting the polynomial, we get –
$ \Rightarrow {x^2} - 5x + 2x - 10$
Now, pulling out like factors from the first two terms in the above equation, we get –
$ \Rightarrow x\left( {x - 5} \right) + 2x - 10$
Taking common factor in the last terms in the above equation, we get –
$ \Rightarrow x\left( {x - 5} \right) + 2\left( {x - 5} \right)$
Hence, the equation becomes - $\left( {x - 5} \right)\left( {x + 2} \right)$
We know that factors of any number are those which when divided that number gives the remainder as 0.
Therefore, in the equation –
$ \Rightarrow \left( {x - 5} \right)\left( {x + 2} \right) = 0$
$
\Rightarrow \left( {x - 5} \right) = 0 \\
\therefore x = 5 \\
$
And –
$
\Rightarrow x + 2 = 0 \\
\therefore x = - 2 \\
$
Hence, this is the required factorization.
Note:
We can also solve this question, by using the Quadratic formula which is –
$x = \dfrac{{ - B \pm \sqrt {{B^2} - 4AC} }}{{2A}}$
Here, $A$ is the coefficient of the first term, $B$ is the coefficient of the second term or middle term and $C$ is the constant term.
Putting the values of $A, B$, and $C$ in the Quadratic formula will give us the exact factorization.
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