How to factorize 25 selected quadratic equations quickly by factor analysis | SureSolv

How to factorize 25 selected quadratic equations quickly by factor analysis

Analyze the factors of the numeric term and the coefficient of $x^2$ together instead of splitting the middle term

how to factorize 25 quadratic equations quickly factor analysis

Factorization of quadratic equations has been an important component of mathematical activities in schools and also in various competitive exams. In school exams or in competitive exams, how quickly and accurately you can factorize a quadratic equation is always a problem we have to grapple with.

Sreedhar Acharya's formula provides a never failing sure-shot method to find the roots of a general quadratic equation. But because of its general nature, in common exams its use remains limited.

Sreedhar Acharya's formula for finding roots of a general quadratic equation

Given a general quadratic equation,

$ax^2+bx+c=0$,

the two roots of $x$ are given by the formula as,

$\displaystyle\frac{-b \pm \sqrt{b^2-4ac}}{2a}$.

To find the roots of the equation for example, $5x^2-87x+378$, you would have to carry out a good amount of calculations. Time being the crucial element in exams, the calculation-heavy Sreedhar Acharya's wonderful formula is not favored. You needed a more practical and quick method.

Middle term splitting method for factorizing a quadratic equation

As a practical method we were taught to split the middle term of a three-term quadratic equation and find the two factors somehow by trial and error.

For example, to find the factors of the quadratic equation,

$x^2-6x+8$,

you would easily split $6=2+4$ so that, third numeric term, $8=2\times{4}$.

You can say with confidence that the two factors of the equation will be,

$x^2-6x+8=(x-2)(x-4)$.

In case of more complex equations, such as $10x^2-x-24=0$, or $5x^2-87x+378=0$, this rudimentary random trial and error method won't serve the purpose of quickly and surely finding the two factors of the quadratic equations.

To meet this need we present a comprehensive method that is systematic and at the same time quick.

How to factorize a quadratic equation quickly and surely—combine analyzing the factors of 1st term coefficient and the 3rd numeric term

Let's take a general quadratic equation to factorize,

$ax^2+bx+c=(a_1x+p)(a_2x+q)=0$.

We have assumed $p$ and $q$ to be the second terms of two factors, and $a_1$ and $a_2$ the coefficients of $x$ in the two factors.

Expanding the RHS and equating the coefficients of like terms and the numeric term we get the all too important relations,

$a_1a_2=a$, the 1st term coefficient,

$pq=c$, the numeric term, and

$a_1q+a_2p=b$, the middle term coefficient.

Whatever be the method you use for factorization of a quadratic equation, if these three relations are satisfied, then only you know that your factorization is correct.

In the comprehensive method, we will mainly analyze how the two factors of the coefficient of the 1st term involving $x^2$, if it is not 1, are multiplied into two products with the two factors of the numeric term to form the middle term coefficient to satisfy,

$a_1q+a_2p=b$.

We'll form feasible factor pairs for each considering the value of the middle term, to quickly form the right combination.

Approach here is from source factors to the destination of the middle term coefficient, not the other way round.

First we will factorize a few example equations to highlight how the method works, and then state the method in an algorithm-like form.

To further consolidate its use, you will solve an exercise of factorizing 25 quadratic equations. Solution to the exercise will conclude this session.

Problem example 1.

Factorize $x^2+17x+42=0$.

Solution example 1.

First step pattern analysis: coefficient of $x^2$ is 1 and so we need to concentrate only on $p$ and $q$, the factors of the numeric term 42.

Second step pattern analysis: middle term and numeric term both are positive, and so $p$ and $q$ both will be positive.

Third step factor analysis: This is the actual step in which we will consider factor pairs 6, 7 and 3, 14 of 42 for their sum to be equal to the middle term coefficient. It is easy to choose 3, 14 so that,

$3+14=17$, the middle term coefficient, and the two factors are,

$x^2+17x+42=(x+3)(x+14)=0$.

Answer. $(x+3)(x+14)$.

If the middle term is changed to 43, the equation to be factorized becomes,

$x^2+43x+42=0$.

In this case, as the middle term coefficient value of 43 is close to, and actually more than the numeric term 42, we will straightaway decide the factor pair of numeric term to be 1, 42.

Problem example 2.

Factorize $2x^2+45+63=0$.

Solution example 2.

Pattern analysis 1: Because of the largeness of the middle term coefficient we will first test feasible factors 21, 3 of the 3rd numeric term. The factors of 1st term are 2, 1. We can easily form the sum 45 as 21 times 2 plus 3 times 1.

Conclusion: Solution comes easily with the required sum as,

$2\times{21}+1\times{3}=45$.

To form the factors of the equation, 2 is to be taken as coefficient of $x$ for 1st factor, and 1 for the second factor. The second numeric term $p$ for the first factor (paired with $2x$) will be 3, the factor of numeric term not multiplied with 2, the coefficient of $x$ of the first term. By the same logic, 21 will be the second numeric term $q$ of the second factor and it will be paired with $x$.

The two factors of the equation are,

$2x^2+45x+63=(2x+3)(x+21)$.

Answer. $(2x+3)(x+21)$.

Problem example 3

Factorize $4x^2-13x-35=0$.

Solution example 3

Pattern analysis 1: The sign of the numeric term is negative and so, the second terms of the two factors of the equation, $p$ and $q$ will be of opposite signs.

Pattern analysis 2: The middle term coefficient will be the difference between the two products of factors of the coefficient of the first term and the numeric term.

Pattern analysis 3: The feasible factors of numeric term 35 are 5, 7 and of the coefficient of the first term, 4, 1 and 2, 2.

Conclusion: 4 times 5 minus 7 is 13, and so the feasible difference of products of factor pairs of 1st term and numeric term is,

$4\times{(-5)}+1\times{7}=-13$.

From this difference of products it is easy to form the factors of the equation,

$4x^2-13x-35=(4x+7)(x-5)$.

Answer. $(4x+7)(x-5)$.

We'll now briefly state the method before taking up consolidating this skill through solving of 25 selected quadratic equations.

Comprehensive method for quick and correct factorization of quadratic equations

First step pattern analysis and actions

If 3rd numeric term is positive, both second terms $p$ and $q$ will be of same sign, and the middle term coefficient will be the sum of the two products between the factors of 1st term coefficient and the factors of the numeric term.

If the numeric term is negative, its two factors are of opposite signs and the middle term coefficient will be formed as a difference of two products between two sets of factors.

Second step pattern analysis and actions

Select feasible factor pairs of the 1st term coefficient and the numeric term.

Ignore extreme end factor pairs if the middle term coefficient value is not very close to the product of 1st term coefficient and the numeric term.

Select first the factor pairs with smaller number of feasible factors, and then select the feasible factor pairs of the other term in consideration of the first set of factors as well as the middle term coefficient value.

Third step of forming the feasible sum of two products of two sets of factors

Explore and form the feasible sum (or difference) of products of two pairs of factors equal to the middle term coefficient as,

$a_1q+a_2p=b$.

If you fail, consider factor pairs ignored in the second step and repeat third step.

Final step of forming the factors of the quadratic equation

From the feasible sum of $a_1q+a_2p=b$, where quadratic equation is $ax^2+bx+c=(a_1+p)(a_2+q)=0$, form the factors,

$(a_1+p)(a_2+q)$.

To clarify and substantiate the method, we have selected 25 quadratic equations for you to factorize and then go through the solutions after the list of questions.

Factorize 25 quadratic equations: Recommended time 12 mins

Problem 1.

Factorize $3x^2-13x+14=0$.

Problem 2.

Factorize $6x^2+77x+121=0$.

Problem 3.

Factorize $x^2-6x-7=0$.

Problem 4.

Factorize $x^2+30x+81=0$.

Problem 5.

Factorize $2x^2+15x+28=0$.

Problem 6.

Factorize $8x^2+18x+9=0$.

Problem 7.

Factorize $6x^2+19x+15=0$.

Problem 8.

Factorize $x^2-3x-88=0$.

Problem 9.

Factorize $2x^2-x-10=0$.

Problem 10.

Factorize $x^2-5x-24=0$.

Problem 11.

Factorize $4x^2-8x+3=0$.

Problem 12.

Factorize $10x^2-7x+1=0$.

Problem 13.

Factorize $10x^2+29x+18=0$.

Problem 14.

Factorize $2x^2+23x+63=0$.

Problem 15.

Factorize $12x^2+19x+5=0$.

Problem 16.

Factorize $9x^2-37x+30=0$.

Problem 17.

Factorize $8x^2+25x+3=0$.

Problem 18.

Factorize $2x^2-21x+40=0$.

Problem 19.

Factorize $28x^2-53y+24=0$.

Problem 20.

Factorize $2x^2+45x+63=0$.

Problem 21.

Factorize $3x^2-49x+200=0$.

Problem 22.

Factorize $5x^2-87x+378=0$.

Problem 23.

Factorize $10x^2-x-24=0$.

Problem 24.

Factorize $3x^2+29x+56=0$.

Problem 25.

Factorize $4x^2-20x+21=0$.


Solutions to factorization of 25 selected quadratic equations

Problem 1.

Factorize $3x^2-13x+14=0$.

Solution 1.

Pattern analysis 1: 1st term factor pair is, 3, 1 and the factor pairs of numeric term are 7, 2 and 14, 1. Ignore 14, 1 as 14 is larger than middle term 13. As middle term is negative, both factors of numeric term are negative.

Conclusion: Feasible sum of products of two pairs of factors is,

$3\times{(-2)}+1\times{(-7)}=-13$.

The two factors of the equation are,

$3x^2-13x+14=(3x-7)(x-2)$.

Answer. $(3x-7)(x-2)$.

Comment: It is just not the factorization of the numeric term into 7 and 2, but also pairing the right factor with the right $x$ term in the two equation factors, that is important.

Problem 2.

Factorize $6x^2+77x+121=0$.

Solution 2.

Pattern analysis 1: Numeric term factor pair is, 11, 11: $121=11\times{11}$, while 1st term coefficient factor pairs are, 3, 2 and 6, 1.

Pattern analysis 2 and conclusion: The numeric term factors are both 11, and the middle term is 7 times 11, so 7 being 6 plus 1, it is straightforward to form the feasible sum of products as,

$6\times{11}+1\times{11}=77$.

The two factors are,

$x^2+77x+121=(6x+11)(x+11)$.

Answer. $(6x+11)(x+11)$.

Problem 3.

Factorize $x^2-6x-7=0$.

Solution 3.

First pattern: 3rd numeric term is negative and so the two factors of numeric term will be of opposite signs, that is, the two partners of $x$ in the two factors of the equation will be of opposite signs.

Second pattern: As $x^2$ coefficient is 1, we have to analyze only the factors of numeric term 7, that is, 7 and 1.

Conclusion: The feasible difference of two products of factors of coefficient of 1st term and numeric term is,

$1\times{1}+1\times{(-7)}=-6$

The two factors of the equation are,

$x^2-6x-7=(x-7)(x+1)$.

Answer. $(x-7)(x+1)$.

Problem 4.

Factorize $x^2+30x+81=0$.

Solution 4.

First pattern: Feasible factor pairs of 81 are, 9, 9 and 27, 3.

Conclusion: The feasible sum of products of factor pairs of coefficient of 1st term and the numeric term is,

$1\times{3}+1\times{27}=30$.

Note: We have considered middle term coefficient value 30 while selecting the final factor pair of the numeric term. This is an essential activity in the method.

The two factors of the equation are,

$x^2+30x+81=(x+27)(x+3)$.

Answer. $(x+27)(x+3)$.

Problem 5.

Factorize $2x^2+15x+28=0$.

Solution 5.

First pattern: 1st term factor pair is 2, 1 and the feasible factor pair of numeric term is 4, 7. The factor pair of 14, 2 is ignored at this stage as 14 is very near to the middle term value of 15.

Conclusion: The feasible sum of two products of 1st term coefficient factors and numeric term factors is,

$2\times{4}+1\times{7}=15$.

The factor 4 will partner $x$ and 7 will partner $2x$. This opposite factor pairing will make the required sum of product equal to 15, coefficient of middle term. Here, the signs of middle term and numeric term being both positive, no complication is created in forming the middle term from the sum of two products.

The two factors of the equation are,

$2x^2+15x+28=(2x+7)(x+4)$.

Answer. $(2x+7)(x+4)$.

Problem 6.

Factorize $8x^2+18x+9=0$.

Solution 6.

First pattern: Factor pairs of numeric term 9 are, 3, 3 and 9, 1. The 1st term coefficient being 8, and middle term coefficient 18, we will ignore at this stage the factor pair 9, 1 of the numeric term as being too large. The feasible factor pair of numeric term at this stage is then 3, 3.

Second pattern: By the same reasoning we will ignore factor pair of 8, 1 of the coefficient 8 of 1st term and take the feasible factor pair as, 4, 2.

Conclusion: The feasible sum of products of the factor pairs of 1st term and numeric term is,

$4\times{3}+2\times{3}=18$ satisfies the middle term coefficient value.

Signs of second and third term being both positive, the two factors are,

$8x^2+18x+9=(4x+3)(2x+3)$.

Answer. $(4x+3)(2x+3)$.

Problem 7.

Factorize $6x^2+19x+15=0$.

Solution 7.

First pattern: Feasible factor pair of numeric term is 5, 3 as the other possibility 15, 1 is ignored as 15 is very close to middle term coefficient value of 19.

Second pattern: Feasible factor pair of 6, the coefficient of $x^2$ is, 3, 2, (again 6, 1 is ignored at this stage).

Conclusion: Feasible sum of products of two sets of factors is,

$3\times{3}+2\times{5}=19$, the coefficient of middle term.

So the two factors are,

$6x^2+19x+15=(3x+5)(2x+3)$.

Answer. $(3x+5)(2x+3)$.

Problem 8.

Factorize $x^2-3x-88=0$.

Solution 8.

First pattern: Sign of numeric 3rd term being negative its two factors are of opposite signs.

Second pattern: Feasible factor pair of 3rd numeric term is 11, 8 as their difference is 3, the value of coefficient of middle term.

Conclusion: The feasible difference of two products of two factor pairs of coefficient of 1st term (1, 1) and numeric term is,

$1\times{8}+1\times{(-11)}=-3$, the value of coefficient of middle term.

The two factors of the equation are,

$x^2-3x-88=(x-11)(x+8)$.

Answer. $(x-11)(x+8)$.

Problem 9.

Factorize $2x^2-x-10=0$.

Solution 9.

First pattern: As numeric 3rd term is negative, its two factors are of opposite signs.

Second pattern: The feasible factor pair of the coefficient of $x^2$ is 2, 1.

Third pattern: Feasible factor pair of 3rd numeric term is then 5, 2. To satisfy the middle term value of -1, the factor -5 is to be paired with $2x$ and factor 2 with $x$.

Conclusion: The sum of the product of two sets of factors is then,

$2\times{2}+1\times{(-5)}=-1$,

So the two factors are,

$2x^2-x-10=(2x-5)(x+2)$.

Answer. $(2x-5)(x+2)$.

Problem 10.

Factorize $x^2-5x-24=0$.

Solution 10.

First pattern: Numeric 3rd term being negative its two factors are of opposite signs.

Second pattern: Considering the middle term coefficient value of -5, the feasible factor pair of numeric term is 8, 3 with difference 5.

Conclusion: The feasible sum of products of two pairs of factors is,

$1\times{3}+1\times{(-8)}=-5$.

The two factors of the equation are,

$x^2-5x-24=(x-8)(x+3)$.

Answer. $(x-8)(x+3)$.

Problem 11.

Factorize $4x^2-8x+3=0$.

Solution 11.

First pattern: Feasible factor pair of numeric term is unique and so is chosen first as, 3, 1. This determines the feasible factor pair for the coefficient of the 1st term 4 as 2, 2 so that 3 times 2 plus 1 times 2 equals 8, the coefficient value of middle term.

Conclusion: The feasible sum of products of two factor pairs is,

$2\times{(-1)}+2\times{(-3)}=-8$.

The two factors of the equation are,

$4x^2-8x+3=(2x-3)(2x-1)$.

Answer. $(2x-3)(2x-1)$.

Problem 12.

Factorize $10x^2-7x+1=0$.

Solution 12.

First pattern: Feasible factor pair of 3rd numeric term is 1, 1 and that of coefficient of 1st term 10 is 5, 2 which satisfies the value of coefficient of middle term.

Conclusion: The feasible sum of products of two factor pairs is,

$5\times{-1}+2\times{-1}=-7$.

The factors of the equation are,

$10x^2-7x+1=(5x-1)(2x-1)$.

Answer. $(5x-1)(2x-1)$.

Problem 13.

Factorize $10x^2+29x+18$.

Solution 13.

First pattern: Feasible factor pairs of 3rd numeric term are, 2, 9 and 3, 6. The factor pair of 18, 1 is ignored as 18 is too close to 29, the middle term coefficient value, especially with 1st term coefficient being also quite large as 10.

Second pattern: The goal of 29 as sum of two products of factors and also considering the two feasible pairs of factors of the numeric term, it is easy to form the only feasible factor pair of coefficient of 1st term as 10, 1. 10 times 2 plus 1 times 9 is 29.

Conclusion: The feasible sum of products of two sets of factors is,

$10\times{2}+1\times{9}=29$.

The two factors are,

$10x^2+29x+18=(10x+9)(x+2)$.

Answer. $(10x+9)(x+2)$. 

Problem 14.

Factorize $2x^2+23x+63=0$.

Solution 14.

First pattern: Unique feasible factor pair of coefficient of 1st term is 2, 1.

Second pattern: With goal 23 of coefficient of middle term as sum of two products of two sets of factors and unique factor pair of coefficient of 1st term, the only feasible factor pair of 3rd term is easy to decide as, 7, 9. The sum of 2 times 7 plus 1 times 9 is 23.

Conclusion: The required sum of products of two factor pairs matching middle term coefficient 23 is,

$2\times{7}+1\times{9}=23$.

The factors of the equation are,

$2x^2+23x+63=(2x+9)(x+7)$.

Answer. $(2x+9)(x+7)$.

Problem 15.

Factorize $12x^2+19x+5=0$.

Solution 15.

First pattern: The unique feasible factor pair of 3rd numeric term is 5, 1.

Second pattern: With the goal of 19 as sum of products of two sets of factors and considering the unique factor pair 5, 1 of 3rd term, it is easy to decide on the only feasible factor pair of coefficient 12 of 1st term as, 3, 4. The sum of 3 times 5 plus 4 times 1 equals 19, the middle term coefficient.

Conclusion: The sum of products of two factor pairs matching the middle term coefficient of 19 is,

$3\times{5}+4\times{1}=19$.

The factors of the equation are,

$12x^2+19x+5=(3x+1)(4x+5)$.

Answer. $(3x+1)(4x+5)$.

Problem 16.

Factorize $9x^2-37x+30=0$.

Solution 16.

First pattern: This is an interesting problem where we will use factors multiples concept. If we split the 1st term as 3, 3, the sum of products of two factor pairs has to be a multiple of 3. As the coefficient of middle term 37 is not a factor of 3, only 9, 1 factor pair of 1st term is valid.

Second pattern: This result decides the only feasible factor pair of numeric term as 3, 10. The sum of 9 times 3 plus 1 times 10 is 37, the value of coefficient of middle term.

Conclusion: The feasible sum of products of two factor pairs is,

$9\times{(-3)}+1\times{(-10)}=-37$.

The two factors of the equation are,

$9x^2-37x+30=(9x-10)(x-3)$.

Answer. $(9x-10)(x-3)$.

Problem 17.

Factorize $8x^2+25x+3=0$.

Solution 17.

Conclusion by analysis of large middle term: The feasible factor pairs of 1st term and the 3rd term can easily be identified as, 8, 1 and 3, 1 so that the sum of products of two factor pairs becomes,

$8\times{3}+1\times{1}=25$.

The two factors of the equation are,

$8x^2+25x+3=(8x+1)(x+3)$.

Answer. $(8x+1)(x+3)$.

Problem 18.

Factorize $2x^2-21x+40=0$.

Solution 18.

First pattern: The 1st factor of factor pair 20, 2 of 3rd term would be too near the middle term coefficient 21 and so ignored. Feasible factor pair of 3rd term are 4, 10 and 8, 5 with only feasible factor pair of 1st term as 2, 1.

Conclusion: Now it is easy to see that 4, 10 factor pair would be invalid and the feasible sum of products of two factor pairs will be,

$2\times{(-8)}+1\times{(-5)}=-21$.

The two factors of the equation are,

$2x^2-21x+40=(2x-5)(x-8)$.

Answer. $(2x-5)(x-8)$.

Problem 19.

Factorize $28x^2-53y+24=0$.

Solution 19.

First pattern: Feasible factor pairs of 1st term are 4, 7 and 14, 2 while that of 3rd term are, 6, 4 and 8, 3. The larger factor pair of 12, 2 is ignored as of now because of large 1st term coefficient of 28. In fact, at this stage, for the same reason, factor pair of 14, 2 can also be ignored.

Conclusion: Testing out the sum of products, the factor pair of 4, 7 and of 8, 3 produces the desired sum,

$4\times{(-8)}+7\times{(-3)}=-53$.

The two factors of the equation are,

$28x^2-53x+24=(4x-3)(7x-8)$.

Answer. $(4x-3)(7x-8)$.

Problem 20.

Factorize $2x^2+45x+63=0$.

Solution 20.

The problem is similar to the problem 14 with both 1st and 3rd terms as same. The only difference is, in this problem, middle term coefficient 45 is very large and quite near to 63.

First pattern: Because of the largeness of the middle term coefficient we will first test feasible factor pair 21, 3 of the 3rd term. The factors of 1st term are 2, 1.

Conclusion: Solution comes easily with the required sum as,

$2\times{21}+1\times{3}$.

The two factors of the equation are,

$2x^2+45x+63=(2x+3)(x+21)$.

Answer. $(2x+3)(x+21)$.

Problem 21.

Factorize $3x^2-49x+200=0$.

Solution 21.

First pattern: Middle term coefficient 49 being close to 40 of 40, 5 factor pair of the large numeric term 200, the feasible factor pairs of 200 are taken as, 10, 20, and 8, 25. The first term coefficient factors are 3 and 1.

Conclusion: It is easy to choose the factor pair 8, 25 giving the desired sum of products of two factor pairs as,

$3\times{(-8)}+1\times{(-25)}=-49$.

The two factors of the equation are,

$3x^2-49x+200=(3x-25)(x-8)$.

Answer. $(3x-25)(x-8)$.

Problem 22.

Factorize $5x^2-87x+378=0$.

Solution 22.

First pattern: 378 is a multiple of 9 and its factors are, 9, 6 and 7. It is good to have an idea of the low valued factors (not too large in number, 3 is suitable) of a large numeric term or coefficient. The factor pair of the coefficient of the 1st term is 5, 1. This information should be enough to quickly choose the right set of factors of 378.

Conclusion: 45 plus 42 is 87, and so the desired sum of products of two sets of factors is,

$5\times{(-9)}+1\times{(-42)}=-87$.

The two factors of the equation are,

$5x^2-87x+378=(5x-42)(x-9)$.

Answer. $(5x-42)(x-9)$.

Problem 23.

Factorize $10x^2-x-24=0$.

Solution 23.

First pattern: The difference between the two products of factor pairs would be 1 as the middle term is -1 and the third term is negative.

Second pattern: The feasible factor pairs of the first term are, 10, 1 and 5, 2. We couldn't ignore 10, 1 as difference in products is concerned. We have no clear idea of the largeness that should be the solution.

Third pattern: The feasible factor pairs of 3rd numeric term is 8, 3 as we have matched the products as 5 times 3 as15 and 2 times 8 as 16 while forming the factor pair of the numeric term.

Conclusion: The desired sum of products of two sets of factors is,

$5\times{3}+2\times{(-8)}=-1$.

The two factors of the equation are,

$10x^2-x-24=(5x-8)(2x+3)$.

Answer. $(5x-8)(2x+3)$.

Problem 24.

Factorize $3x^2+29x+56=0$.

Solution 24.

First pattern: Unique factor pair of coefficient of 1st term is 3, 1. As middle term coefficient 29 is very near to 28 of the factor pair of 28, 2 of numeric term, the factor pair is ignored. The feasible factor pairs of numeric term 56 are, 14, 4 and 8, 7.

Conclusion: The sum $21+8=29$, or, $3\times{7}+1\times{8}=29$ satisfies the middle term requirement.

The two factors of the equation are,

$3x^2+29x+56=(3x+8)(x+7)$.

Answer. $(3x+8)(x+7)$.

Problem 25.

Factorize $4x^2-20x+21=0$.

Solution 25.

First pattern: 21 being too close to 20, the factor pair 21, 1 of numeric term is ignored. The feasible factor pair of the numeric term is, 3, 7.

Second pattern and conclusion: As middle term coefficient is a multiple of 2 and sum of two factors of numeric term is 10, the desired sum of two pairs of factors is,

$2\times{(-3)}+2\times{(-7)}=-20$.

The two factors of the equation are,

$4x^2-20+21=(2x-7)(2x-3)$.

Answer. $(2x-7)(2x-3)$.

The more you practice, the more concise and quick your reasoning will be.