
How do you factor the expression \[21{{x}^{2}}+29x-10\]?
Answer
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Hint: From the question given, we have been asked to factor the expression \[21{{x}^{2}}+29x-10\]. We can factor the given expression by using the factorization method. We know that factorization is the process of finding the factors for the given expression. First of all, we have to split the given expression into the simplified form in which we can get the factors for the given expression.
Complete step by step answer:
From the question given, we have been given the expression \[21{{x}^{2}}+29x-10\]
First of all, as of process, we have to multiply the coefficient of \[{{x}^{2}}\] and constant. Then, we have to write the obtained value as the product of two numbers. \[-210=35\times -6\]
By multiplying the coefficient of \[{{x}^{2}}\] and constant and then by writing the obtained number as the product of the two numbers, we get
\[21{{x}^{2}}+29x-10\]
\[\Rightarrow 21{{x}^{2}}+35x-6x-10\]
Now, we have to take the common terms out and then we have to further simplify the expression.
By taking the terms common out and then further simplifying the expression, we get
\[\Rightarrow 7x\left( 3x+5 \right)-2\left( 3x+5 \right)\]
\[\Rightarrow \left( 3x+5 \right)\left( 7x-2 \right)\]
Above two are the factors for the given expression in the question.
Therefore, we got the factors for the given expression by using the process which we use for factoring.
Note: We should be well known about the process of factorization. By using this process, we can get the factors for the given expression very easily. Also, we should be very careful while factorizing the given expression. Also, we should be very careful while simplifying the expression. Similarly we can find the solutions by first obtaining the solutions of the given quadratic equation $a{{x}^{2}}+bx+c=0$ using the formulae $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ and for the given equation \[21{{x}^{2}}+29x-10\] as $x=\dfrac{-29\pm \sqrt{{{29}^{2}}-4\left( 21 \right)\left( -10 \right)}}{2\left( 21 \right)}=\dfrac{-5}{3},\dfrac{2}{7}$ .
Complete step by step answer:
From the question given, we have been given the expression \[21{{x}^{2}}+29x-10\]
First of all, as of process, we have to multiply the coefficient of \[{{x}^{2}}\] and constant. Then, we have to write the obtained value as the product of two numbers. \[-210=35\times -6\]
By multiplying the coefficient of \[{{x}^{2}}\] and constant and then by writing the obtained number as the product of the two numbers, we get
\[21{{x}^{2}}+29x-10\]
\[\Rightarrow 21{{x}^{2}}+35x-6x-10\]
Now, we have to take the common terms out and then we have to further simplify the expression.
By taking the terms common out and then further simplifying the expression, we get
\[\Rightarrow 7x\left( 3x+5 \right)-2\left( 3x+5 \right)\]
\[\Rightarrow \left( 3x+5 \right)\left( 7x-2 \right)\]
Above two are the factors for the given expression in the question.
Therefore, we got the factors for the given expression by using the process which we use for factoring.
Note: We should be well known about the process of factorization. By using this process, we can get the factors for the given expression very easily. Also, we should be very careful while factorizing the given expression. Also, we should be very careful while simplifying the expression. Similarly we can find the solutions by first obtaining the solutions of the given quadratic equation $a{{x}^{2}}+bx+c=0$ using the formulae $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ and for the given equation \[21{{x}^{2}}+29x-10\] as $x=\dfrac{-29\pm \sqrt{{{29}^{2}}-4\left( 21 \right)\left( -10 \right)}}{2\left( 21 \right)}=\dfrac{-5}{3},\dfrac{2}{7}$ .
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