
How do you factor completely \[2{x^2} + 7x - 15 = 0\]?
Answer
492.3k+ views
Hint:In order to determine the factors of the above quadratic question use the Splitting up the middle method by first multiplying the coefficient of${x^2}$ with the constant term and factorise it into two factors such that either addition or subtraction gives us the middle term and the product of the same gives us back the multiplication we have calculated. Pull out common from the first two terms and the last two terms and then again pulling out the binomial parenthesis will give you the required factorisation.
Complete step by step solution:
Given a quadratic equation, \[2{x^2} + 7x - 15 = 0\] let it be $f(x)$
$f(x) = 2{x^2} + 7x - 15 = 0$
Comparing the equation with the standard Quadratic equation $a{x^2} + bx + c$
a becomes 2
b becomes 7
And c becomes 15
To find the quadratic factorization we’ll use splitting up the middle term method
So first calculate the product of coefficient of ${x^2}$ and the constant term which comes to be $ = ( - 15) \times 2 = - 30$
Now the second Step is to find the 2 factors of the number 2 such that the whether addition or subtraction of those numbers is equal to the middle term or coefficient of x and the product of those factors result in the value of constant .
So if we factorize -30, the answer comes to be 10 and -3 as $10 - 3 = 7$ that is the middle term . and
\[ - 3 \times 10 = - 30\] which is perfectly equal to the constant value.
Now writing the middle term sum of the factors obtained ,so equation $f(x)$ becomes
$f(x) = 2{x^2} + 10x - 3x - 15 = 0$
Now taking common from the first 2 terms and last 2 terms
$f(x) = 2x(x + 5) - 3(x + 5)$
Finding the common binomial parenthesis, the equation becomes
$f(x) = (x + 5)(2x - 3)$
Hence , We have successfully factorized our quadratic equation.
Therefore the factors are$(x + 5)$ and $(2x - 3)$ .
Alternative:
You can also alternatively use a direct method which uses Quadratic Formula to find both roots of a quadratic equation as
$x1 = \dfrac{{ - b + \sqrt {{b^2} - 4ac} }}{{2a}}$ and $x2 = \dfrac{{ - b - \sqrt {{b^2} - 4ac} }}{{2a}}$
x1,x2 are root to quadratic equation $a{x^2} + bx + c$
Hence the factors will be $(x - x1)\, and \,(x - x2)\,$.
Note:
1. One must be careful while calculating the answer as calculation error may come.
2.Don’t forget to compare the given quadratic equation with the standard one every time.
3. Write the factors when the middle term is split using the proper sign.
Complete step by step solution:
Given a quadratic equation, \[2{x^2} + 7x - 15 = 0\] let it be $f(x)$
$f(x) = 2{x^2} + 7x - 15 = 0$
Comparing the equation with the standard Quadratic equation $a{x^2} + bx + c$
a becomes 2
b becomes 7
And c becomes 15
To find the quadratic factorization we’ll use splitting up the middle term method
So first calculate the product of coefficient of ${x^2}$ and the constant term which comes to be $ = ( - 15) \times 2 = - 30$
Now the second Step is to find the 2 factors of the number 2 such that the whether addition or subtraction of those numbers is equal to the middle term or coefficient of x and the product of those factors result in the value of constant .
So if we factorize -30, the answer comes to be 10 and -3 as $10 - 3 = 7$ that is the middle term . and
\[ - 3 \times 10 = - 30\] which is perfectly equal to the constant value.
Now writing the middle term sum of the factors obtained ,so equation $f(x)$ becomes
$f(x) = 2{x^2} + 10x - 3x - 15 = 0$
Now taking common from the first 2 terms and last 2 terms
$f(x) = 2x(x + 5) - 3(x + 5)$
Finding the common binomial parenthesis, the equation becomes
$f(x) = (x + 5)(2x - 3)$
Hence , We have successfully factorized our quadratic equation.
Therefore the factors are$(x + 5)$ and $(2x - 3)$ .
Alternative:
You can also alternatively use a direct method which uses Quadratic Formula to find both roots of a quadratic equation as
$x1 = \dfrac{{ - b + \sqrt {{b^2} - 4ac} }}{{2a}}$ and $x2 = \dfrac{{ - b - \sqrt {{b^2} - 4ac} }}{{2a}}$
x1,x2 are root to quadratic equation $a{x^2} + bx + c$
Hence the factors will be $(x - x1)\, and \,(x - x2)\,$.
Note:
1. One must be careful while calculating the answer as calculation error may come.
2.Don’t forget to compare the given quadratic equation with the standard one every time.
3. Write the factors when the middle term is split using the proper sign.
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