
How do you factor $ 2{x^2} - 3x - 9 $ ?
Answer
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Hint: In this question, we need to factor $ 2{x^2} - 3x - 9 $ . First, we will find two numbers that multiply to give $ ac $ i.e., $ 2 \times - 9 = - 18 $ , and add to give $ b $ i.e., $ - 3 $ which is called a sum-product pattern. Then rewrite the middle with those numbers. Then, factor the first two and last two terms separately. If we have done this correctly, then two new terms will have a clearly visible common factor.
Complete step-by-step answer:
Now, we need to factor $ 2{x^2} - 3x - 9 $ .
According to the rule to factorize,
Product= $ {x^2} $ coefficient $ \times $ constant
And, sum= $ x $ coefficient
Thus, we will find two numbers that multiply to give $ ac $ i.e., $ 2 \times - 9 = - 18 $ and add to give $ b $ i.e., $ - 3 $ ,
Here, the product is negative. So, we can say that one of the factors is negative, and then the other is positive.
Now, let’s consider the possible factors and their sum.
$
18 \times - 1 = - 18;18 + \left( { - 1} \right) = 17 \\
9 \times - 2 = - 18;9 + \left( { - 2} \right) = 7 \\
6 \times - 3 = - 18;6 + \left( { - 3} \right) = 3 \\
3 \times - 6 = - 18;3 + \left( { - 6} \right) = - 3 \\
2 \times - 9 = - 18;2 + \left( { - 9} \right) = - 7 \\
1 \times - 18 = - 18;1 + \left( { - 18} \right) = - 17 \;
$
From this it is clear that the factors are $ 3 $ and $ - 6 $ .
Now, by rewriting the middle term with those factors, we have,
$ = 2{x^2} + 3x - 6x - 9 $
Factor out the greatest common factor from each group,
$ = x\left( {2x + 3} \right) - 3\left( {2x + 3} \right) $
Factor the polynomial by factoring out the greatest common factor, $ 2x + 3 $ ,
= $ \left( {x - 3} \right)\left( {2x + 3} \right) $
Hence, the factors of $ 2{x^2} - 3x - 9 $ is $ \left( {2x + 3} \right) $ and $ \left( {x - 3} \right) $ .
So, the correct answer is “ $ \left( {2x + 3} \right) $ and $ \left( {x - 3} \right) $”.
Note: In this question it is important to note here that this factorization method works for all quadratic equations. The standard form of the quadratic equation is $ a{x^2} + bx + c = 0 $ . It is called factoring because we find the factors. A factor is something we multiply by. There is no simple method of factoring a quadratic expression, but with a little practice it becomes easier. Here, the middle step can be skipped with some practice.
Complete step-by-step answer:
Now, we need to factor $ 2{x^2} - 3x - 9 $ .
According to the rule to factorize,
Product= $ {x^2} $ coefficient $ \times $ constant
And, sum= $ x $ coefficient
Thus, we will find two numbers that multiply to give $ ac $ i.e., $ 2 \times - 9 = - 18 $ and add to give $ b $ i.e., $ - 3 $ ,
Here, the product is negative. So, we can say that one of the factors is negative, and then the other is positive.
Now, let’s consider the possible factors and their sum.
$
18 \times - 1 = - 18;18 + \left( { - 1} \right) = 17 \\
9 \times - 2 = - 18;9 + \left( { - 2} \right) = 7 \\
6 \times - 3 = - 18;6 + \left( { - 3} \right) = 3 \\
3 \times - 6 = - 18;3 + \left( { - 6} \right) = - 3 \\
2 \times - 9 = - 18;2 + \left( { - 9} \right) = - 7 \\
1 \times - 18 = - 18;1 + \left( { - 18} \right) = - 17 \;
$
From this it is clear that the factors are $ 3 $ and $ - 6 $ .
Now, by rewriting the middle term with those factors, we have,
$ = 2{x^2} + 3x - 6x - 9 $
Factor out the greatest common factor from each group,
$ = x\left( {2x + 3} \right) - 3\left( {2x + 3} \right) $
Factor the polynomial by factoring out the greatest common factor, $ 2x + 3 $ ,
= $ \left( {x - 3} \right)\left( {2x + 3} \right) $
Hence, the factors of $ 2{x^2} - 3x - 9 $ is $ \left( {2x + 3} \right) $ and $ \left( {x - 3} \right) $ .
So, the correct answer is “ $ \left( {2x + 3} \right) $ and $ \left( {x - 3} \right) $”.
Note: In this question it is important to note here that this factorization method works for all quadratic equations. The standard form of the quadratic equation is $ a{x^2} + bx + c = 0 $ . It is called factoring because we find the factors. A factor is something we multiply by. There is no simple method of factoring a quadratic expression, but with a little practice it becomes easier. Here, the middle step can be skipped with some practice.
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