How do you find the product $\left( {{x^2} - 4x + 5} \right)\left( {5{x^2} + 3x - 4} \right)$?
Answer
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Hint:In distributive property of multiplication over addition, we can add the numbers and then multiply, or we can multiply and then add. First multiply ${x^2}$ with each term in $\left( {5{x^2} + 3x - 4} \right)$. So, first ${x^2}$ is multiplied with $5{x^2}$, then the product is added to the product of ${x^2}$ and $3x$, ${x^2}$ and $ - 4$. Now, multiply $ - 4x$ with each term in $\left( {5{x^2} + 3x - 4} \right)$. For this, $ - 4x$ is multiplied with $5{x^2}$, then the product is added to the product of $ - 4x$ and $3x$, $ - 4x$ and $ - 4$. Now, multiply $5$ with each term in $\left( {5{x^2} + 3x - 4} \right)$. For this, $5$ is multiplied with $5{x^2}$, then the product is added to the product of $5$ and $3x$, $5$ and $ - 4$. Then, add the simplified version of ${x^2}\left( {5{x^2} + 3x - 4} \right)$, $ - 4x\left( {5{x^2} + 3x - 4} \right)$ and $5\left( {5{x^2} + 3x - 4} \right)$ to get the product of $\left( {{x^2} - 4x + 5} \right)\left( {5{x^2} + 3x - 4} \right)$.
Formula used:
Distributive property of multiplication over addition:
Let $a$, $b$ and $c$ be three real numbers, then
$a \times (b + c) = (a \times b) + (a \times c)$
Complete step by step answer:
To “distributive” means to divide something or give a share or part of something. According to the distributive property, multiplying the difference of two numbers will give the same result as multiplying each addend individually by the number and then subtracting the products.
The distributive property helps in making difficult problems simpler by writing it in simpler form. The distributive property of multiplication is used to rewrite expression by distributing or breaking down a factor as a sum or difference of two numbers.
Given: $\left( {{x^2} - 4x + 5} \right)\left( {5{x^2} + 3x - 4} \right)$
To multiply these two terms, we multiply each individual term in the left parenthesis by each individual term in the right parenthesis.
So, first multiply ${x^2}$ with each term in $\left( {5{x^2} + 3x - 4} \right)$.
We can add the numbers and then multiply, or we can multiply and then add.
Here, ${x^2}\left( {5{x^2} + 3x - 4} \right)$
So, first ${x^2}$ is multiplied with $5{x^2}$, then the product is added to the product of ${x^2}$ and $3x$, ${x^2}$ and $ - 4$.
${x^2}\left( {5{x^2} + 3x - 4} \right) = {x^2}\left( {5{x^2}} \right) + {x^2}\left( {3x} \right) + {x^2}\left( { - 4} \right)$
So, multiplying ${x^2}$ with $5{x^2}$, and multiplying ${x^2}$ with $3x$, and multiplying ${x^2}$ with $ - 4$, we get
$ \Rightarrow {x^2}\left( {5{x^2} + 3x - 4} \right) = 5{x^4} + 3{x^3} - 4{x^2}$
Therefore, simplified version of ${x^2}\left( {5{x^2} + 3x - 4} \right)$ is $5{x^4} + 3{x^3} - 4{x^2}$.
Now, multiply $ - 4x$ with each term in $\left( {5{x^2} + 3x - 4} \right)$.
We can add the numbers and then multiply, or we can multiply and then add.
Here, $ - 4x\left( {5{x^2} + 3x - 4} \right)$
So, first $ - 4x$ is multiplied with $5{x^2}$, then the product is added to the product of $ - 4x$ and $3x$, $ - 4x$ and $ - 4$.
$ - 4x\left( {5{x^2} + 3x - 4} \right) = - 4x\left( {5{x^2}} \right) - 4x\left( {3x} \right) - 4x\left( { - 4} \right)$
So, multiplying $ - 4x$ with $5{x^2}$, and multiplying $ - 4x$ with $3x$, and multiplying $ - 4x$ with $ - 4$, we get
$ \Rightarrow - 4x\left( {5{x^2} + 3x - 4} \right) = - 20{x^3} - 12{x^2} + 16x$
Therefore, the simplified version of $ - 4x\left( {5{x^2} + 3x - 4} \right)$ is $ - 20{x^3} - 12{x^2} + 16x$.
Now, multiply $5$ with each term in $\left( {5{x^2} + 3x - 4} \right)$.
We can add the numbers and then multiply, or we can multiply and then add.
Here, $5\left( {5{x^2} + 3x - 4} \right)$
So, first $5$ is multiplied with $5{x^2}$, then the product is added to the product of $5$ and $3x$, $5$ and $ - 4$.
$5\left( {5{x^2} + 3x - 4} \right) = 5\left( {5{x^2}} \right) + 5\left( {3x} \right) + 5\left( { - 4} \right)$
So, multiplying $5$ with $5{x^2}$, and multiplying $5$ with $3x$, and multiplying $5$ with $ - 4$, we get
$ \Rightarrow 5\left( {5{x^2} + 3x - 4} \right) = 25{x^2} + 15x - 20$
Therefore, the simplified version of $5\left( {5{x^2} + 3x - 4} \right)$ is $25{x^2} + 15x - 20$.
Now, add the simplified version of ${x^2}\left( {5{x^2} + 3x - 4} \right)$, $ - 4x\left( {5{x^2} + 3x - 4} \right)$ and $5\left( {5{x^2} + 3x - 4} \right)$.
Since, simplified version of ${x^2}\left( {5{x^2} + 3x - 4} \right)$ is $5{x^4} + 3{x^3} - 4{x^2}$, and simplified version of $ - 4x\left( {5{x^2} + 3x - 4} \right)$ is $ - 20{x^3} - 12{x^2} + 16x$ , and simplified version of $5\left( {5{x^2} + 3x - 4} \right)$ is $25{x^2} + 15x - 20$.
So, add $5{x^4} + 3{x^3} - 4{x^2}$, $ - 20{x^3} - 12{x^2} + 16x$ and $25{x^2} + 15x - 20$.
${x^2}\left( {5{x^2} + 3x - 4} \right) - 4x\left( {5{x^2} + 3x - 4} \right) + 5\left( {5{x^2} + 3x - 4} \right)$
$ = 5{x^4} + 3{x^3} - 4{x^2} - 20{x^3} - 12{x^2} + 16x + 25{x^2} + 15x - 20$
Add $3{x^3}$ and $ - 20{x^3}$, add $ - 4{x^2}$, $ - 12{x^2}$ and $25{x^2}$, add $16x$ and $15x$, we get
$ \Rightarrow {x^2}\left( {5{x^2} + 3x - 4} \right) - 4x\left( {5{x^2} + 3x - 4} \right) + 5\left( {5{x^2} + 3x - 4} \right)$
$ = 5{x^4} - 17{x^3} + 9{x^2} + 31x - 20$
Final solution: Therefore, the product of $\left( {{x^2} - 4x + 5} \right)\left( {5{x^2} + 3x - 4} \right)$ is $5{x^4} - 17{x^3} + 9{x^2} + 31x - 20$.
Note:
The distributive property of addition and subtraction can be used to rewrite expressions for a variety of purposes. When we are multiplying a number by a sum, we can add and then multiply. We can also multiply each addend first and then add the products. This can be done with subtraction as well, multiplying each number in the difference before subtracting. In this case, we are distributing the outside multiplier to each number in the parentheses, so that multiplication occurs with each number before addition or subtraction occurs.
Even though division is the inverse of multiplication, the distributive law only holds true in case of division, when the dividend is distributed or broken down. For instance, using the distributive law for $132 \times 6$, $132$ can be broken down as $60 + 60 + 12$, thus making division easier. However, $132 \times \left( {4 + 2} \right)$ will give you the wrong result.
Formula used:
Distributive property of multiplication over addition:
Let $a$, $b$ and $c$ be three real numbers, then
$a \times (b + c) = (a \times b) + (a \times c)$
Complete step by step answer:
To “distributive” means to divide something or give a share or part of something. According to the distributive property, multiplying the difference of two numbers will give the same result as multiplying each addend individually by the number and then subtracting the products.
The distributive property helps in making difficult problems simpler by writing it in simpler form. The distributive property of multiplication is used to rewrite expression by distributing or breaking down a factor as a sum or difference of two numbers.
Given: $\left( {{x^2} - 4x + 5} \right)\left( {5{x^2} + 3x - 4} \right)$
To multiply these two terms, we multiply each individual term in the left parenthesis by each individual term in the right parenthesis.
So, first multiply ${x^2}$ with each term in $\left( {5{x^2} + 3x - 4} \right)$.
We can add the numbers and then multiply, or we can multiply and then add.
Here, ${x^2}\left( {5{x^2} + 3x - 4} \right)$
So, first ${x^2}$ is multiplied with $5{x^2}$, then the product is added to the product of ${x^2}$ and $3x$, ${x^2}$ and $ - 4$.
${x^2}\left( {5{x^2} + 3x - 4} \right) = {x^2}\left( {5{x^2}} \right) + {x^2}\left( {3x} \right) + {x^2}\left( { - 4} \right)$
So, multiplying ${x^2}$ with $5{x^2}$, and multiplying ${x^2}$ with $3x$, and multiplying ${x^2}$ with $ - 4$, we get
$ \Rightarrow {x^2}\left( {5{x^2} + 3x - 4} \right) = 5{x^4} + 3{x^3} - 4{x^2}$
Therefore, simplified version of ${x^2}\left( {5{x^2} + 3x - 4} \right)$ is $5{x^4} + 3{x^3} - 4{x^2}$.
Now, multiply $ - 4x$ with each term in $\left( {5{x^2} + 3x - 4} \right)$.
We can add the numbers and then multiply, or we can multiply and then add.
Here, $ - 4x\left( {5{x^2} + 3x - 4} \right)$
So, first $ - 4x$ is multiplied with $5{x^2}$, then the product is added to the product of $ - 4x$ and $3x$, $ - 4x$ and $ - 4$.
$ - 4x\left( {5{x^2} + 3x - 4} \right) = - 4x\left( {5{x^2}} \right) - 4x\left( {3x} \right) - 4x\left( { - 4} \right)$
So, multiplying $ - 4x$ with $5{x^2}$, and multiplying $ - 4x$ with $3x$, and multiplying $ - 4x$ with $ - 4$, we get
$ \Rightarrow - 4x\left( {5{x^2} + 3x - 4} \right) = - 20{x^3} - 12{x^2} + 16x$
Therefore, the simplified version of $ - 4x\left( {5{x^2} + 3x - 4} \right)$ is $ - 20{x^3} - 12{x^2} + 16x$.
Now, multiply $5$ with each term in $\left( {5{x^2} + 3x - 4} \right)$.
We can add the numbers and then multiply, or we can multiply and then add.
Here, $5\left( {5{x^2} + 3x - 4} \right)$
So, first $5$ is multiplied with $5{x^2}$, then the product is added to the product of $5$ and $3x$, $5$ and $ - 4$.
$5\left( {5{x^2} + 3x - 4} \right) = 5\left( {5{x^2}} \right) + 5\left( {3x} \right) + 5\left( { - 4} \right)$
So, multiplying $5$ with $5{x^2}$, and multiplying $5$ with $3x$, and multiplying $5$ with $ - 4$, we get
$ \Rightarrow 5\left( {5{x^2} + 3x - 4} \right) = 25{x^2} + 15x - 20$
Therefore, the simplified version of $5\left( {5{x^2} + 3x - 4} \right)$ is $25{x^2} + 15x - 20$.
Now, add the simplified version of ${x^2}\left( {5{x^2} + 3x - 4} \right)$, $ - 4x\left( {5{x^2} + 3x - 4} \right)$ and $5\left( {5{x^2} + 3x - 4} \right)$.
Since, simplified version of ${x^2}\left( {5{x^2} + 3x - 4} \right)$ is $5{x^4} + 3{x^3} - 4{x^2}$, and simplified version of $ - 4x\left( {5{x^2} + 3x - 4} \right)$ is $ - 20{x^3} - 12{x^2} + 16x$ , and simplified version of $5\left( {5{x^2} + 3x - 4} \right)$ is $25{x^2} + 15x - 20$.
So, add $5{x^4} + 3{x^3} - 4{x^2}$, $ - 20{x^3} - 12{x^2} + 16x$ and $25{x^2} + 15x - 20$.
${x^2}\left( {5{x^2} + 3x - 4} \right) - 4x\left( {5{x^2} + 3x - 4} \right) + 5\left( {5{x^2} + 3x - 4} \right)$
$ = 5{x^4} + 3{x^3} - 4{x^2} - 20{x^3} - 12{x^2} + 16x + 25{x^2} + 15x - 20$
Add $3{x^3}$ and $ - 20{x^3}$, add $ - 4{x^2}$, $ - 12{x^2}$ and $25{x^2}$, add $16x$ and $15x$, we get
$ \Rightarrow {x^2}\left( {5{x^2} + 3x - 4} \right) - 4x\left( {5{x^2} + 3x - 4} \right) + 5\left( {5{x^2} + 3x - 4} \right)$
$ = 5{x^4} - 17{x^3} + 9{x^2} + 31x - 20$
Final solution: Therefore, the product of $\left( {{x^2} - 4x + 5} \right)\left( {5{x^2} + 3x - 4} \right)$ is $5{x^4} - 17{x^3} + 9{x^2} + 31x - 20$.
Note:
The distributive property of addition and subtraction can be used to rewrite expressions for a variety of purposes. When we are multiplying a number by a sum, we can add and then multiply. We can also multiply each addend first and then add the products. This can be done with subtraction as well, multiplying each number in the difference before subtracting. In this case, we are distributing the outside multiplier to each number in the parentheses, so that multiplication occurs with each number before addition or subtraction occurs.
Even though division is the inverse of multiplication, the distributive law only holds true in case of division, when the dividend is distributed or broken down. For instance, using the distributive law for $132 \times 6$, $132$ can be broken down as $60 + 60 + 12$, thus making division easier. However, $132 \times \left( {4 + 2} \right)$ will give you the wrong result.
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