
Factorise the given equation : \[7{p^2} + 49pq\]
Answer
565.2k+ views
Hint: Here, we are required to factorise the given polynomial. We will use the method of ‘Completing the square’ to expand the given polynomial. Then using the algebraic identity we will simplify the equation further to find the required factors.
Formula Used:
We will use the following formulas:
1.\[{a^2} + 2ab + {b^2} = {\left( {a + b} \right)^2}\]
2.\[{a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right)\]
Complete step-by-step answer:
The given quadratic polynomial is: \[7{p^2} + 49pq\]
We will use the method of ‘Completing the square’ to expand this polynomial.
This means that we have to change this polynomial in the form of \[{a^2} + 2ab + {b^2}\] by adding and subtracting terms such that the given polynomial remains the same.
Hence, by completing the square, we can write the given polynomial as:
\[7{p^2} + 49pq = {\left( {\sqrt 7 p} \right)^2} + 2\left( {\sqrt 7 p} \right)\left( {\dfrac{{7\sqrt 7 q}}{2}} \right) + {\left( {\dfrac{{7\sqrt 7 q}}{2}} \right)^2} - {\left( {\dfrac{{7\sqrt 7 q}}{2}} \right)^2}\]
If we observe this carefully, then LHS is equal to RHS just the way of writing the polynomial has changed.
Now, since, the polynomial is in the form of \[{a^2} + 2ab + {b^2}\], we will use the formula \[{a^2} + 2ab + {b^2} = {\left( {a + b} \right)^2}\]to substitute it.
\[ \Rightarrow 7{p^2} + 49pq = {\left( {\sqrt 7 p + \dfrac{{7\sqrt 7 q}}{2}} \right)^2} - {\left( {\dfrac{{7\sqrt 7 q}}{2}} \right)^2}\]
Now, using the formula \[{a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right)\], we can write the above RHS as:
\[ \Rightarrow 7{p^2} + 49pq = \left( {\sqrt 7 p + \dfrac{{7\sqrt 7 q}}{2} + \dfrac{{7\sqrt 7 q}}{2}} \right)\left( {\sqrt 7 p + \dfrac{{7\sqrt 7 q}}{2} - \dfrac{{7\sqrt 7 q}}{2}} \right)\]
\[ \Rightarrow 7{p^2} + 49pq = \left( {\sqrt 7 p + 7\sqrt 7 q} \right)\left( {\sqrt 7 p} \right)\]
Taking \[\sqrt 7 \]common from the first bracket in RHS, we get
\[ \Rightarrow 7{p^2} + 49pq = \left( {p + 7q} \right)\left( {\sqrt 7 \times \sqrt 7 p} \right)\]
\[ \Rightarrow 7{p^2} + 49pq = \left( {p + 7q} \right)\left( {7p} \right)\]
Hence, this is the required answer.
Note: An alternate way to solve this question is that instead of using a complicated method like ‘Completing the square’, we can also use the direct method of greatest common factor.
In this method, we directly take out the greatest factor which is common from the polynomial and leave the rest of the terms inside the bracket. The bracket becomes one factor and the common terms taken out become the other factor of the given polynomial.
Hence, given polynomial is: \[7{p^2} + 49pq\]
The greatest common factor is \[7p\].
Hence, taking it out from the bracket, we get,
\[7{p^2} + 49pq = 7p\left( {p + 7q} \right)\]
Hence, this is the required answer.
Formula Used:
We will use the following formulas:
1.\[{a^2} + 2ab + {b^2} = {\left( {a + b} \right)^2}\]
2.\[{a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right)\]
Complete step-by-step answer:
The given quadratic polynomial is: \[7{p^2} + 49pq\]
We will use the method of ‘Completing the square’ to expand this polynomial.
This means that we have to change this polynomial in the form of \[{a^2} + 2ab + {b^2}\] by adding and subtracting terms such that the given polynomial remains the same.
Hence, by completing the square, we can write the given polynomial as:
\[7{p^2} + 49pq = {\left( {\sqrt 7 p} \right)^2} + 2\left( {\sqrt 7 p} \right)\left( {\dfrac{{7\sqrt 7 q}}{2}} \right) + {\left( {\dfrac{{7\sqrt 7 q}}{2}} \right)^2} - {\left( {\dfrac{{7\sqrt 7 q}}{2}} \right)^2}\]
If we observe this carefully, then LHS is equal to RHS just the way of writing the polynomial has changed.
Now, since, the polynomial is in the form of \[{a^2} + 2ab + {b^2}\], we will use the formula \[{a^2} + 2ab + {b^2} = {\left( {a + b} \right)^2}\]to substitute it.
\[ \Rightarrow 7{p^2} + 49pq = {\left( {\sqrt 7 p + \dfrac{{7\sqrt 7 q}}{2}} \right)^2} - {\left( {\dfrac{{7\sqrt 7 q}}{2}} \right)^2}\]
Now, using the formula \[{a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right)\], we can write the above RHS as:
\[ \Rightarrow 7{p^2} + 49pq = \left( {\sqrt 7 p + \dfrac{{7\sqrt 7 q}}{2} + \dfrac{{7\sqrt 7 q}}{2}} \right)\left( {\sqrt 7 p + \dfrac{{7\sqrt 7 q}}{2} - \dfrac{{7\sqrt 7 q}}{2}} \right)\]
\[ \Rightarrow 7{p^2} + 49pq = \left( {\sqrt 7 p + 7\sqrt 7 q} \right)\left( {\sqrt 7 p} \right)\]
Taking \[\sqrt 7 \]common from the first bracket in RHS, we get
\[ \Rightarrow 7{p^2} + 49pq = \left( {p + 7q} \right)\left( {\sqrt 7 \times \sqrt 7 p} \right)\]
\[ \Rightarrow 7{p^2} + 49pq = \left( {p + 7q} \right)\left( {7p} \right)\]
Hence, this is the required answer.
Note: An alternate way to solve this question is that instead of using a complicated method like ‘Completing the square’, we can also use the direct method of greatest common factor.
In this method, we directly take out the greatest factor which is common from the polynomial and leave the rest of the terms inside the bracket. The bracket becomes one factor and the common terms taken out become the other factor of the given polynomial.
Hence, given polynomial is: \[7{p^2} + 49pq\]
The greatest common factor is \[7p\].
Hence, taking it out from the bracket, we get,
\[7{p^2} + 49pq = 7p\left( {p + 7q} \right)\]
Hence, this is the required answer.
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