12a^3bc-18a^2bc^2+6a^2c, x^2+20x+36, 12x^4-8x^3-4x^2, x^2+36x, 9x^2+6x+1, 6x^3+11x-10, x^2+3x-6..............can someone please help me solve some of these problems if not all. I have forgotten how to solve them can someone please help..Thanks
I'd be happy to help you solve these problems step-by-step. Let's go through each one of them:
1. 12a^3bc - 18a^2bc^2 + 6a^2c:
Step 1: First, let's factor out the common factors from each term.
12a^3bc - 18a^2bc^2 + 6a^2c = 6a^2c(2ab - 3bc + 1)
2. x^2 + 20x + 36:
Step 1: This is a quadratic equation, so let's try factoring it.
x^2 + 20x + 36 = (x + 6)(x + 6) = (x + 6)^2
3. 12x^4 - 8x^3 - 4x^2:
Step 1: Let's factor out the common factor, which is 4x^2.
12x^4 - 8x^3 - 4x^2 = 4x^2(3x^2 - 2x - 1)
Step 2: Now, we'll try to factor the quadratic expression inside the parentheses.
4x^2(3x^2 - 2x - 1) = 4x^2(3x + 1)(x - 1)
4. x^2 + 36x:
Step 1: Let's factor out the common factor, which is x.
x^2 + 36x = x(x + 36)
5. 9x^2 + 6x + 1:
Step 1: This is a quadratic expression and cannot be easily factored. We can use the quadratic formula to find the roots.
The quadratic formula is x = (-b ± √(b^2 - 4ac))/(2a)
For this equation, a = 9, b = 6, and c = 1.
Plugging in the values, we get x = (-6 ± √(6^2 - 4*9*1))/(2*9)
Simplifying, we get x = (-6 ± √(36 - 36))/(18)
Since the discriminant is zero, the equation has a single real root: x = -6/18 = -1/3
6. 6x^3 + 11x - 10:
Step 1: This is another quadratic expression that cannot be factored easily. We can again use the quadratic formula to find the roots.
The quadratic formula is x = (-b ± √(b^2 - 4ac))/(2a)
For this equation, a = 6, b = 11, and c = -10.
Plugging in the values, we get x = (-11 ± √(11^2 - 4*6*(-10)))/(2*6)
Simplifying, we get x = (-11 ± √(121 + 240))/(12)
x = (-11 ± √(361))/(12)
Simplifying further, x = (-11 ± 19)/(12)
The two roots are x = 8/6 = 4/3 and x = -30/12 = -5/2.
7. x^2 + 3x - 6:
Step 1: We'll try to factor the quadratic expression.
x^2 + 3x - 6 = (x - 1)(x + 6)
I hope this helps! Let me know if you have any further questions.
Of course! I'll be happy to help you solve these problems. Let's go one by one:
1. 12a^3bc - 18a^2bc^2 + 6a^2c:
To simplify this expression, we can factor out the greatest common factor (GCF), which in this case is 6ac:
6ac(2a^2b - 3abc + c)
2. x^2 + 20x + 36:
This is a quadratic trinomial in the form of ax^2 + bx + c. To solve it, we can try factoring or using the quadratic formula. In this case, it can be factored as:
(x + 18)(x + 2)
3. 12x^4 - 8x^3 - 4x^2:
This is also a quadratic trinomial, but it's in terms of x^4, x^3, and x^2. We can factor out the GCF, which is 4x^2:
4x^2(3x^2 - 2x - 1)
4. x^2 + 36x:
Here we have a quadratic binomial. We can factor out the greatest common factor, which is x:
x(x + 36)
5. 9x^2 + 6x + 1:
This is a quadratic trinomial. To solve it, we can factor or use the quadratic formula. In this case, it cannot be easily factored, so let's use the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / (2a)
By substituting the values of a, b, and c into the formula, we get:
x = (-6 ± √(6^2 - 4(9)(1))) / (2(9))
Simplifying the equation, we have:
x = (-6 ± √(36 - 36)) / 18
x = (-6 ± √(0)) / 18
x = -6 / 18 or x = -6 / 18
So, the only solution is:
x = -1/3
6. 6x^3 + 11x - 10:
This is a cubic polynomial. Unfortunately, it cannot be easily factored, so we can use different methods for finding the solution, such as synthetic division or numerical methods like Newton's method. For simplicity, let's use a numerical method or a calculator to approximate the solution, which is x ≈ -1.17253.
7. x^2 + 3x - 6:
This is a quadratic trinomial. Similar to previous examples, we can try factoring or use the quadratic formula. In this case, it cannot be easily factored, so let's use the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / (2a)
By substituting the values of a, b, and c into the formula, we get:
x = (-3 ± √(3^2 - 4(1)(-6))) / (2(1))
Simplifying the equation, we have:
x = (-3 ± √(9 + 24)) / 2
x = (-3 ± √33) / 2
Since the square root of 33 cannot be simplified further, we can keep the expression as is.
I hope this helps! If you have any additional questions or need further assistance, please let me know.