Question

How do you factor the trinomial \[4{x^2} - 21x + 5 = 0\] ?

Answer 1

Hint: To solve the given equation by factoring, find two numbers that multiply to give a times c and add to give b of the quadratic equation of the form \[a{x^2} + bx + c\] , in which we need to factor by grouping and combine all the like terms, in which we can easily find the factors of the equation using AC method.

Complete step-by-step answer:
Let us write the given equation
 \[4{x^2} - 21x + 5 = 0\]
We can consider that the equation is of the form \[a{x^2} + bx + c\] , by which we can easily find the factors of the equation using the AC method.
Using the AC method of factorising a quadratic, consider the factors of \[\left( {4 \times 5} \right)\] which sum to - 21 the factors of \[ + 20\] which sum to - 21 are - 20 and - 1 splitting the middle term gives:
 \[4{x^2} - 20x - x + 5\] i.e., factor by grouping
= \[4x\left( {x - 5} \right) - 1\left( {x - 5} \right)\]
Now let us take out the common factor and simplify all terms, we get
 \[\left( {x - 5} \right)\left( {4x - 1} \right)\]
 \[ \Rightarrow 4{x^2} - 21x + 5 = \left( {x - 5} \right)\left( {4x - 1} \right)\]
Hence, the factors are
 \[\left( {x - 5} \right)\left( {4x - 1} \right)\]
And to solve the equation:
 \[\left( {x - 5} \right)\left( {4x - 1} \right) = 0\]
Equate each of the factors to zero and solve for x i.e.,
 \[\left( {x - 5} \right) = 0\]
 \[\left( {4x - 1} \right) = 0\]
Now let us solve for the first factor i.e.,
 \[x - 5 = 0\]
Therefore, we get
 \[x = 5\]
Now let us solve for the second factor i.e.,
 \[4x - 1 = 0\]
Therefore, we get
 \[x = \dfrac{1}{4}\]

Note: The key point to factor the trinomial equation is to group the factors using factoring method i.e., of the form \[a{x^2} + bx + c\] , in this given quadratic equation we need to find two integers whose product is equal to c and the sum is equal to b using AC method. Then solve each factor obtained by setting it to zero by this we can get the value of b of both the factors.