
How do you solve $4{x^2} + 28x - 32 = 0$ by factoring?
Answer
570k+ views
Hint: In the method of factoring, taking the reference of the equation $a{x^2} + bx + c = 0$, we have to find any two numbers whose multiplication results in a product of term $a$ and $c$ . Also, the sum of two numbers, if the term $c$ is positive and the difference between two numbers, if the term $c$ is negative should result in the term $b$.
Complete step-by-step answer:
First of all, we compare the given equation $4{x^2} + 28x - 32 = 0$ with the standard equation $a{x^2} + bx + c = 0$ .
From the comparison, we get
$\Rightarrow$$a = 4$
$\Rightarrow$$b = 28$
$\Rightarrow$$c = - 32$
Here, the term $c$ is negative.
Hence, we need to find two numbers whose difference will lead to $b$ and their product will lead to a product of term $a$ and $c$ .
As per the problem, we need two numbers that follow the conditions
Product = $a \times c = 4 \times 32 = 128$
Difference = $b = 28$ .
For the product $\;128$ , the pairs can be $\;1$ and $\;128$ , $\;2$ and $\;64$ , $\;4$ and $\;32$ , $\;8$ and $\;16$
Out of these all pairs, the pair $\;4$ and $\;32$ seems to be valid because
Product = $4 \times 32 = 128 = a \times c$
Difference = $32 - 4 = 28 = b$
Now, we have two numbers for factoring.
Now, consider the equation
$\Rightarrow$$4{x^2} + 28x - 32 = 0$
Now, we can replace the term $\;28$ with the difference
$ \Rightarrow 4{x^2} + (32 - 4)x - 32 = 0$
Opening the brackets,
$ \Rightarrow 4{x^2} + 32x - 4x - 32 = 0$
Shifting the terms,
$ \Rightarrow 4{x^2} - 4x + 32x - 32 = 0$
Now, considering the first two terms and expanding them
$ \Rightarrow 4 \times x \times x - 4 \times x$
Here, in both terms $\;4$ and $x$ are common
So taking $\;4$ and $x$ common and putting the remaining in the brackets
$ \Rightarrow 4x(x - 1)$
Similarly, for the last two terms
$ \Rightarrow 32 \times x - 32$
Here, in both terms $\;32$ is common
So taking $\;32$ common and putting the remaining in the brackets
$ \Rightarrow 32(x - 1)$
Therefore, replacing the equations we get,
$ \Rightarrow 4x(x - 1) + 32(x - 1) = 0$
From this equation, taking the bracket $(x - 1)$ common, we get
$ \Rightarrow (x - 1)(4x + 32) = 0$
Here either of the two brackets is equal to zero.
$ \Rightarrow (x - 1) = 0$
$ \Rightarrow x = 1$
Also considering the other bracket
$ \Rightarrow (4x + 32) = 0$
Shifting $\;32$ on the right-hand side
$ \Rightarrow 4x = - 32$
$ \Rightarrow x = - \dfrac{{32}}{4}$
Hence the final answer can be written as
$ \Rightarrow x = - 8$
Hence, we have two values for $x$
$x = 1$ or $x = - 8$
Note:
Here, the conditions to remember for the two numbers depends on the sign of $c$. If $c$ is positive, we need to take the sum of two numbers to give the value $b$. If $c$ is negative, we need to take the difference of two numbers to give the value $b$ . The alternate formula to find the value of $x$ is $\dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$.
Complete step-by-step answer:
First of all, we compare the given equation $4{x^2} + 28x - 32 = 0$ with the standard equation $a{x^2} + bx + c = 0$ .
From the comparison, we get
$\Rightarrow$$a = 4$
$\Rightarrow$$b = 28$
$\Rightarrow$$c = - 32$
Here, the term $c$ is negative.
Hence, we need to find two numbers whose difference will lead to $b$ and their product will lead to a product of term $a$ and $c$ .
As per the problem, we need two numbers that follow the conditions
Product = $a \times c = 4 \times 32 = 128$
Difference = $b = 28$ .
For the product $\;128$ , the pairs can be $\;1$ and $\;128$ , $\;2$ and $\;64$ , $\;4$ and $\;32$ , $\;8$ and $\;16$
Out of these all pairs, the pair $\;4$ and $\;32$ seems to be valid because
Product = $4 \times 32 = 128 = a \times c$
Difference = $32 - 4 = 28 = b$
Now, we have two numbers for factoring.
Now, consider the equation
$\Rightarrow$$4{x^2} + 28x - 32 = 0$
Now, we can replace the term $\;28$ with the difference
$ \Rightarrow 4{x^2} + (32 - 4)x - 32 = 0$
Opening the brackets,
$ \Rightarrow 4{x^2} + 32x - 4x - 32 = 0$
Shifting the terms,
$ \Rightarrow 4{x^2} - 4x + 32x - 32 = 0$
Now, considering the first two terms and expanding them
$ \Rightarrow 4 \times x \times x - 4 \times x$
Here, in both terms $\;4$ and $x$ are common
So taking $\;4$ and $x$ common and putting the remaining in the brackets
$ \Rightarrow 4x(x - 1)$
Similarly, for the last two terms
$ \Rightarrow 32 \times x - 32$
Here, in both terms $\;32$ is common
So taking $\;32$ common and putting the remaining in the brackets
$ \Rightarrow 32(x - 1)$
Therefore, replacing the equations we get,
$ \Rightarrow 4x(x - 1) + 32(x - 1) = 0$
From this equation, taking the bracket $(x - 1)$ common, we get
$ \Rightarrow (x - 1)(4x + 32) = 0$
Here either of the two brackets is equal to zero.
$ \Rightarrow (x - 1) = 0$
$ \Rightarrow x = 1$
Also considering the other bracket
$ \Rightarrow (4x + 32) = 0$
Shifting $\;32$ on the right-hand side
$ \Rightarrow 4x = - 32$
$ \Rightarrow x = - \dfrac{{32}}{4}$
Hence the final answer can be written as
$ \Rightarrow x = - 8$
Hence, we have two values for $x$
$x = 1$ or $x = - 8$
Note:
Here, the conditions to remember for the two numbers depends on the sign of $c$. If $c$ is positive, we need to take the sum of two numbers to give the value $b$. If $c$ is negative, we need to take the difference of two numbers to give the value $b$ . The alternate formula to find the value of $x$ is $\dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$.
Recently Updated Pages
Master Class 11 Business Studies: Engaging Questions & Answers for Success

Master Class 11 Computer Science: Engaging Questions & Answers for Success

Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 Social Science: Engaging Questions & Answers for Success

Master Class 11 English: Engaging Questions & Answers for Success

Master Class 11 Chemistry: Engaging Questions & Answers for Success

Trending doubts
Draw a diagram of nephron and explain its structur class 11 biology CBSE

Explain zero factorial class 11 maths CBSE

Chemical formula of Bleaching powder is A Ca2OCl2 B class 11 chemistry CBSE

Name the part of the brain responsible for the precision class 11 biology CBSE

The growth of tendril in pea plants is due to AEffect class 11 biology CBSE

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

