The given polynomial does not have factors with rational coefficients.
Oh, dude, it's like this: all quadratic equations are polynomials, but not all polynomials are quadratic equations. A quadratic equation is a specific type of polynomial that has a degree of 2, meaning it has a highest power of x^2. So, like, all squares are rectangles, but not all rectangles are squares, you know what I mean?
No, it’s true. It’s the same as saying if 60 is divided by 2 and the remainder equals zero (no remainder, so it divides perfectly), 2 is a factor of 60.
x3 - 2x2 + x - 2 =(x - 2)(x2 + 1)
True Yes. Although the term 'quad' stands for four, a quadratic equation is a polynomial of second degree.
(x + 2)(x - 8)
12a2-8a-2 2(6a2-4a-1)
The GCF of 6a^2 and 8a is 2a.
When we solve an equation in mathematics we say that we find its root. Let f(x) = 0 be an equation. A root of the equation is a value k such that f(k) = 0. If f(x) is a polynomial, then f(x) = 0 is a polynomial equation. By the Factor Theorem, k is a root of this equation if and only if (x - k) is a factor of f(x). If (x - k) is a factor of f(x), then k is a simple root. If (x - k)^2 is a factor of f(x), then k is a double root. If (x - k)^3 is a factor of f(x), then k is a triple root, and so on. Thus, we can say that a root of order n, where n = 2 or n > 2, is a multiple (or repeated) root.
(x-2)(x-3)
-3b2(5b - 2)
2(2g2 - 25)
You can factor a polynomial using one of these steps: 1. Factor out the greatest common monomial factor. 2. Look for a difference of two squares or a perfect square trinomial. 3. Factor polynomials in the form ax^2+bx+c into a product of binomials. 4. Factor a polynomial with 4 terms by grouping.
Suppose you have a polynomial, p(x) = a0 + a1x + a2x^2 + a3x^3 + ... + anx^n then (ax - b) is a factor of the polynomial if and only if p(b/a) = 0
False
(x-2)(x+2)
15x^2+3x-12 3(5x^2+x-4)=Answer
x^2 + 5x - 24 = (x - 3)(x + 8)