A quadratic polynomial is of the form of
ax^{2} + bx + c = 0
where a, b, c are real numbers.
How to factor quadratic expressions ?
Step 1 :
Write the product of the coefficient of the first and last term with their signs and the product of the factors equal to that particular value and when we simplify those two we should get the middle term.
Step 2 :
Replace the middle term by the sum of factors.
Step 3 :
Using grouping method, find the factors.
Problem 1 :
2x^{2} – 9x - 5
Solution :
Step 1 :
Product of factors = -10
Sum of factors = -9
Step 2 :
(-5) · 2 = -10 and (-5) + 2 = -3 ≠ -9
1 · (-10) = -10 and 1 + (-10) = -9
So, the factors are -10 and 1.
Step 3 :
= 2x^{2} + 1x – 10x – 5
By grouping,
= (2x^{2} + 1x) + (-10x – 5)
Step 4 :
By taking the common factor, we get
= x(2x + 1) - 5(2x + 1)
= (x – 5)(2x + 1)
Problem 2 :
3x^{2} + 5x - 2
Solution :
Step 1 :
Product of factors = -6
Sum of factors = 5
Step 2 :
(-2) · 3 = -6 and (-2) + 3 = 1 ≠ 5
(-1) · 6 = -6 and (-1) + 6 = 5
So, the factors are -1 and 6.
Step 3 :
= 3x^{2} - 1x + 6x – 2
By grouping,
= (3x^{2} - 1x) + (6x – 2)
Step 4 :
By taking the common factor, we get
= x(3x – 1) + 2(3x - 1)
= (x + 2)(3x – 1)
Problem 3 :
3x^{2} – 5x - 2
Solution :
Step 1 :
Product of factors = -6
Sum of factors = -5
Step 2 :
(-2) · 3 = -6 and (-2) + 3 = 1 ≠ -5
1 · (-6) = -6 and 1 + (-6) = -5
So, the factors are 1 and -6.
Step 3 :
= 3x^{2} + 1x - 6x – 2
By grouping,
= (3x^{2} + 1x) + (-6x - 2)
Step 4 :
By taking the common factor, we get
= x(3x + 1) - 2(3x + 1)
= (x - 2)(3x + 1)
Problem 4 :
2x^{2} + 3x - 2
Solution :
Step 1 :
Product of factors = -4
Sum of factors = 3
Step 2 :
Finding the factors of product and sum.
(-2) · 2 = -4 and (-2) + 2 = 0 ≠ 3
(-1) · 4 = -4 and (-1) + 4 = 3
So, the factors are -1 and 4.
Step 3 :
= 2x^{2} - 1x + 4x – 2
By grouping,
= (2x^{2} - 1x) + (4x – 2)
Step 4 :
By taking the common factor, we get
= x(2x – 1) + 2(2x - 1)
= (x + 2)(2x – 1)
Problem 5 :
2x^{2} + 3x - 5
Solution :
Step 1 :
Product of factors = -10
Sum of factors = 3
Step 2 :
(-1) · 10 = -10 and (-1) + 10 = 9 ≠ 3
(-2) · 5 = -10 and (-2) + 5 = 3
So, the factors are -2 and 5.
Step 3 :
= 2x^{2} - 2x + 5x – 5
By grouping,
= (2x^{2} - 2x) + (5x – 5)
Step 4 :
By taking the common factor, we get
= 2x(x – 1) + 5(x - 1)
= (x - 1)(2x + 5)
Problem 6 :
5x^{2} – 14x - 3
Solution :
Step 1 :
Product of factors = -15
Sum of factors = -14
Step 2 :
(-3) · 5 = -15 and (-3) + 5 = 2 ≠ -14
1 · (-15) = -15 and 1 + (-15) = -14
So, the factors are 1 and -15.
Step 3 :
= 5x^{2} + x - 15x - 3
By grouping,
= (5x^{2} + x) + (-15x - 3)
Step 4 :
By taking the common factor, we get
= x(5x + 1) - 3(5x + 1)
= (x - 3)(5x + 1)
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