How do you factor $3{x^2} + 17x + 10$?

Hint:Factorization refers to writing a mathematical term as a product of several smaller terms of the same kind. These smaller terms are known as factors. For factoring a quadratic function, we can split the coefficient of $x$ such that its product is equal to the product of the coefficient of ${x^2}$ and the constant. Then, we can take the common factors from the first two terms and the last two terms. After that we can simplify it to get the required factors.

Complete step by step solution:
(i)We have the quadratic equation:
$3{x^2} + 17x + 10$

Let us assume the given quadratic equation as $Q$,
$Q = 3{x^2} + 17x + 10$

As we know that, to factorize the quadratic equation of the form $a{x^2} + bx + c$, we can split the co- efficient of $x$ i.e., $b$ as the sum of two numbers such that their product is the product of the co- efficient of ${x^2}$ and the constant i.e., $a$ and $c$. Let say, we split $b$ into $n1$ and $n2$, then the two conditions should be:

$n1 + n2 = b$
And,
$n1 \times n2 = ac$

Now, as we have $3{x^2} + 17x + 10$ as our function, we will compare it with $a{x^2} + bx + c$ to
obtain the values of $a$, $b$ and $c$. Therefore, after comparing, we will get:
$
a = 3 \\
b = 17 \\
c = 10 \\
$

(ii)Since, we now know the values of $a$, $b$ and $c$, we will try to think of two numbers whose sum is

$17$ and their product is $30$, i.e.,
$n1 + n2 = 17$
And,
$n1 \times n2 = 30$

(iii)After thinking about these two conditions, by hit and trial method we see that the sum of $15$ and $2$
is $17$ and their product is $30$, i.e.,

$15 + 2 = 17$
And,
$15 \times 2 = 30$
Therefore, we can split our middle term as $15x$ and $2x$.

(iv)So, after splitting $17x$ as $15x + 2x$, our quadratic equation will become:
$Q = 3{x^2} + 15x + 2x + 10$

Now, we need to take the common factor from the first two terms and the last two terms. So, as we can see $3x$ is common in the first two terms and $2$ is common in the last two terms, we will eventually get:
$Q = 3x(x + 5) + 2(x + 5)$

Now, we take the common factor $(x + 5)$ from both the terms, we get:
$Q = (x + 5)(3x + 2)$

Hence, the quadratic expression can be factored as $(x + 5)(3x + 2)$.

Note: If you are facing difficulty while thinking about the two numbers in which the middle term has to be split by hit and trial method, you can also directly find the roots of the given quadratic equation through the formula $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$ and then let say the two obtained roots are $x1$ and $x2$, you can directly write the factored form of the quadratic equation as $(x - x1)(x - x2)$.