from the sum of 11 x cube minus 3 y square + 8 and 2 x cube + 12 y square + 1 subtract the sum of 9 x cube + y square - 1 and 3 x cube + 3 y + 18
Answers
Step-by-step explanation:
Tutorial with detailled solutions on factoring polynomials.
Factoring Formulas
1: One common factor. a x + a y = a (x + y)
2: Sevearl grouped common factor. a x + a y + b x + b y = a(x + y) + b(x + y) = (a + b ) (x + y)
3: Difference of two squares (1). x 2 - y 2 = (x + y)(x - y)
4: Difference of two squares (2). (x + y) 2 - z 2 = (x + y + z)(x + y - z)
5: Sum of two cubes. x 3 + y 3 = (x + y)(x 2 - x y + y 2)
6: Difference of two cubes. x 3 - y 3 = (x - y)(x 2 + x y + y 2)
7: Difference of fourth powers. x 4 - y 4 = (x 2 - y 2)(x 2 + y 2) = (x + y)(x - y)(x 2 + y 2)
8: Perfet square x 2 + 2xy + y 2 = (x + y) 2
9: Perfet square x 2 - 2xy + y 2 = (x - y) 2
10: Perfect cube x 3 + 3x 2y + 3xy 2 + y 3 = (x + y) 3
11: Perfect cube x 3 - 3x 2y + 3xy 2 - y 3 = (x - y) 3
Example 1: Factor the binomial 9 - 4x 2
Solution
Rewrite the given expression as the difference of two squares then apply formula 1 given above.
9 - 4x 2 = 3 2 - (2x) 2 = (3 - 2x)(3 + 2x)
As a practice, multiply (3 - 2x)(3 + 2x) to obtain the given expression.
Example 2: Factor the trinomial 9x 2 + 3x - 2
Solution
To factor the above trinomial, we need to write it in the form.
9x 2 + 3x - 2 = (ax + m)(bx + n)
Expand the product on the right above
9x 2 + 3x - 2 = abx 2 + x(mb + na) + mn
For the polynomial on the left to be equal to the polynomial on the right we need to have equal corresponding coefficients, hence
ab = 9
mb + na = 3
mn = -2
Trial values for a and b are: a = 1 and b = 9 or a = 3 and b = 3
Trial values for m and n are: m = 1 and n = -2, m = 2 and n = -1, m = -1 and n = 2 and m = -2 and n = 1.
Trying various values for a, b, m and n among the list above, we arrive at:
9x 2 + 3x - 2 = (3x + 2)(3x - 1)
As a practice, multiply (3x + 2)(3x - 1) and simplify to obtain the given trinomial.
Example 3: Factor the polynomial x 3 + 2x 2 - 16x - 32
Solution
Group terms that have common factors.
x 3 + 2x 2 - 16x - 32 = (x 3 + 2x 2) - (16x + 32)
Factor the grouped terms
= x 2(x + 2) - 16(x + 2)
Factor x + 2 out
= (x + 2)(x 2 - 16)
The term (x 2 - 16) is the difference of two squares and can be factored using formula 1 above
= (x + 2)(x + 4)(x - 4)
Check the above answer by expanding the obtained result.
Exercises: Factor the polynomials.
1: (x + 1) 2 - 4
2: x 2 + 5x + 6
3: x 3 - 1
4: x 3 - x 2 - 25x + 25
Solutions to above exercises
1: (x + 1) 2 - 4 = (x - 1)(x + 3)
2: x 2 + 5x + 6 = (x + 2)(x + 3)
3: x 3 - 1 = (x - 1)(x 2 + x + 1)
4: x 3 - x 2 - 25x + 25 = (x - 1)(x + 5)(x - 5)
More references and links to polynomial functions.
Multiplicity of Zeros and Graphs Polynomials.
Find Zeros of Polynomial Functions - Problems
Polynomial Functions, Zeros, Factors and Intercepts
Step-by-step explanation:
answer of this question is 4.