Question

The equation $3{x^2} - 17x + 20$ can be factored as:
A.$\left( {x + 4} \right)\left( {3x + 5} \right)$
B.$\left( {x - 4} \right)\left( {3x - 5} \right)$
C.$\left( {x - 4} \right)\left( {x - 5} \right)$
D.None of these

Answer 1

Hint: For factorising, 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 1st two terms and the last two terms. After that we can simplify it to get the required factors.

Complete step-by-step answer:
We have the quadratic equation $3{x^2} - 17x + 20$.
Let $Q = 3{x^2} - 17x + 20$
To factorise the quadratic equation of the form $a{x^2} + bx + c$, we can write b as the sum of 2 numbers which give the product of a and c.
Here we have $b = - 17$. We can write it as a sum of $ - 12$ and $ - 5$. We get the product of $ - 12$ and $ - 5$ as 60. We also have the product of a and c as $3 \times 20 = 60$
So we can write the equation as
\[ \Rightarrow \]$Q = 3{x^2} - 12x - 5x + 20$
We can take the common factor $3x$ from the 1st 2 terms and 5 from the last 2 terms. So we get,
\[ \Rightarrow \]$Q = 3x\left( {x - 4} \right) - 5\left( {x - 4} \right)$
We can take the common factor $\left( {x - 4} \right)$ from both the terms, so we get,
\[ \Rightarrow \]$ Q = \left( {x - 4} \right)\left( {3x - 5} \right)$
Therefore, the quadratic expression can be factored as $\left( {x - 4} \right)\left( {3x - 5} \right)$
So, the correct answer is option B.


Note: Alternate approach to the problem is by substituting $x = 1$ in the given expression and in the options and comparing their values.
For $x = 1$, the expression will become,
$3{x^2} - 17x + 20 = 3 - 17 + 20 = 6$
For $x = 1$, option A will become,
$\left( {x + 4} \right)\left( {3x + 5} \right) = 5 \times 8 = 40 \ne 6$
So option A is not a factor.
For$x = 1$, option B will become,
$\left( {x - 4} \right)\left( {3x - 5} \right) = - 3 \times - 2 = 6$

So option B is a factor.
For$x = 1$, option C will become,
$\left( {x - 4} \right)\left( {x - 5} \right) = - 3 \times - 4 = 12 \ne 6$
So option C is not a factor.
Therefore, the quadratic expression can be factored as $\left( {x - 4} \right)\left( {3x - 5} \right)$