How do you factor and solve ${x^2} - 8x + 15 = 0$?
Answer
589.5k+ 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, ${x^2} - 8x + 15 = 0$ let it be $f(x)$
$f(x) = {x^2} - 8x + 15 = 0$
Comparing the equation with the standard Quadratic equation $a{x^2} + bx + c$
a becomes 1
b becomes -8
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 1 = 15$
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 15 ,the answer comes to be -5 and -3 as $ - 5 - 3 = - 8$ that is the middle term and $ - 5 \times - 3 = 15$ which is perfectly equal to the constant value.
Now writing the middle term sum of the factors obtained ,so equation $f(x)$ becomes
$f(x) = {x^2} - 5x - 3x + 15$
Now taking common from the first 2 terms and last 2 terms
$f(x) = x(x - 5) - 3(x - 5)$
Finding the common binomial parenthesis, the equation becomes
$
\Rightarrow f(x) = (x - 5)(x - 3) = 0 \\
\Rightarrow x - 5 = 0 \\
\Rightarrow x = 5 \\
\Rightarrow x - 3 = 0 \\
\Rightarrow x = 3 \\
$
Hence , We have successfully factorized our quadratic equation.
Therefore the factors are $(x - 3)$ and $(x - 5)$ and solution is $x = 5, 3$
Alternative:
You can also alternatively use a direct method which uses Quadratic Formula to find both roots of a quadratic equation as
$x_1 = \dfrac{{ - b + \sqrt {{b^2} - 4ac} }}{{2a}}$ and $x_2 = \dfrac{{ - b - \sqrt {{b^2} - 4ac} }}{{2a}}$
$x_1,x_2$ are root to quadratic equation $a{x^2} + bx + c$
Hence the factors will be $(x - x_1)\,and\,(x - x_2)\,$.
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, ${x^2} - 8x + 15 = 0$ let it be $f(x)$
$f(x) = {x^2} - 8x + 15 = 0$
Comparing the equation with the standard Quadratic equation $a{x^2} + bx + c$
a becomes 1
b becomes -8
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 1 = 15$
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 15 ,the answer comes to be -5 and -3 as $ - 5 - 3 = - 8$ that is the middle term and $ - 5 \times - 3 = 15$ which is perfectly equal to the constant value.
Now writing the middle term sum of the factors obtained ,so equation $f(x)$ becomes
$f(x) = {x^2} - 5x - 3x + 15$
Now taking common from the first 2 terms and last 2 terms
$f(x) = x(x - 5) - 3(x - 5)$
Finding the common binomial parenthesis, the equation becomes
$
\Rightarrow f(x) = (x - 5)(x - 3) = 0 \\
\Rightarrow x - 5 = 0 \\
\Rightarrow x = 5 \\
\Rightarrow x - 3 = 0 \\
\Rightarrow x = 3 \\
$
Hence , We have successfully factorized our quadratic equation.
Therefore the factors are $(x - 3)$ and $(x - 5)$ and solution is $x = 5, 3$
Alternative:
You can also alternatively use a direct method which uses Quadratic Formula to find both roots of a quadratic equation as
$x_1 = \dfrac{{ - b + \sqrt {{b^2} - 4ac} }}{{2a}}$ and $x_2 = \dfrac{{ - b - \sqrt {{b^2} - 4ac} }}{{2a}}$
$x_1,x_2$ are root to quadratic equation $a{x^2} + bx + c$
Hence the factors will be $(x - x_1)\,and\,(x - x_2)\,$.
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.
Recently Updated Pages
Master Class 11 English: Engaging Questions & Answers for Success

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

Master Class 11 Maths: Engaging Questions & Answers for Success

Master Class 11 Biology: Engaging Questions & Answers for Success

Master Class 11 Physics: Engaging Questions & Answers for Success

Master Class 11 Chemistry: Engaging Questions & Answers for Success

Trending doubts
Explain the Treaty of Vienna of 1815 class 10 social science CBSE

First Cotton Mill of India

In which year voting age was reduced from 21 to 18?

What planets have no moons Which one has only one moon class 10 physics CBSE

What is the full form of POSCO class 10 social science CBSE

10 examples of evaporation in daily life with explanations

