How to factorize x^2 + 5x - 36
How can we factor $$x^2 + 5x - 36$$?
Answer: $$x^2+5x-36=(x+9)(x-4)$$
The factors of 36: 1, 2, 3, 4, 6, 9, 12, 18, and 36.
Find a pair of factors of $$36$$ which differ by $$5$$.
As you can see above, the pair (4, 9) works because $$9$$ x $$4=36$$ and $$9 - 4 =5$$
So, the answer is $$x^2+5x-36=(x+9)(x-4)$$
You can also factorize by solving the quadratic equation
The equation x² + 5x - 36 = 0 has 2 real roots when solved:
x₁ = -9 and x₂ = 4
Thus, $$x^2+5x-36=(x+9)(x-4)$$
Solving the equation
An equation like ax² + bx + c = 0, can be solved by using the quadratic equation formula:
or
Where
See step-by-step solution below:
Identify the coefficients
a = 1, b = 5 and c = -36
Evaluate the value of Delta
Δ = b² - 4ac
Δ = 5² - 4.1.(-36) = 25 - 4.(-36)
Δ = 25 - (-144) = 169
Plug the values of a, b and Δ (the discriminant) into the Bhaskara formula
x = -5 ± √1692.1
x = -5 ± √1692 (general solution)
As we can see above, the discriminant (Δ) of this equation is positive (Δ > 0) meaning that there are two real roots (two solutions), x₁ and x₂.
To find x₁, we just choose the negative sign before the square root of delta. So,
x₁ = -5 - √1692 = -5 - 132 = -182 = -9
To find x₂, we just choose the positive sign before the square root of delta. So,
x₂ = -5 + √1692 = -5 + 132 = 82 = 4
S = {-9, 4}