What is the factor of 16^2-81

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Answer 1

Answer: this is the diffiarence of two perfect squares. this should be memorised, but if not here it is

a^2-b^2=(a-b)(a+b)
so 16^2-81=16^2-9^2
so the answer is (16-9)(16+9)

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The reciprocal of an integer plus the reciprocal of two times the integer plus two equals 2/3 Find the integer.

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$(1)/(x) + (1)/(2x+2)= (2)/(3)\ /\cdot 3x(2x+2) \ \ \ \wedge\ \ \ x \neq 0\ \ \wedge\ \ 2x+2 \neq 0\n \n.\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x\in R-\{0;-1\}\n \n (1)/(x)\cdot 3x(2x+2) + (1)/(2x+2)\cdot 3x(2x+2)= (2)/(3)\cdot 3x(2x+2)\n \n3(2x+2)+3x=2x(2x+2)\n \n6x+6+3x=4x^2+4x\n \n-4x^2+9x-4x+6=0\n \n-4x^2+5x+6=0\ \ \ \Rightarrow\ \ \Delta=5^2-4\cdot(-4)\cdot6=25+96=121\n \n$

$x_1= (-5- √(121) )/(2\cdot (-4))= (-5-11)/(-8) = (-16)/(-8) =2\ \ \ is\ integer\n \nx_2= (-5+ √(121) )/(2\cdot (-4))= (-5+11)/(-8) = (6)/(-8) =- (3)/(4) \ \ \ is\ not\ the\ integer$

According to the Fundamental Theorem of Algebra, how many roots exist for the polynomial function? f(x) = 8x7 – x5 + x3+6

Answers

According to the Fundamental Theorem of Algebra, the number of roots of a polynomial is equal to the degree of the polynomial. The degree of the polynomial is the highest exponent of a term in the polynomial.
Looking at the function, the term with the highest exponent is 8x7. The exponent is 7; therefore, the function has 7 roots.

According to the Fundamental Theorem of Algebra, the roots exist for the polynomial function $f\left( x \right) = 8{x^7} - {x^5} + {x^3} + 6 x$ is $\boxed7.$

Further explanation:

The Fundamental Theorem of Algebra states that the polynomial has n roots if the degree of the polynomial is n.

$f\left( x \right) = a{x^n} + b{x^(n - 1)} +\ldots + cx + d$

The polynomial function has n roots or zeroes.

Degree is highest power of the polynomial function.

Given:

The polynomial function is $f\left( x \right) = 8{x^7} - {x^5} + {x^3} + 6.$

Explanation:

The polynomial function $f\left( x \right) = 8{x^7} - {x^5} + {x^3} + 6$ has seven zeroes as the degree of the polynomial is 7.

According to the Fundamental Theorem of Algebra, the roots exist for the polynomial function $f\left( x \right) = 8{x^7} - {x^5} + {x^3} + 6$ is $\boxed7.$

Learn more:

1. Learn more about inverse of the functionbrainly.com/question/1632445.

2. Learn more about equation of circle brainly.com/question/1506955.

3. Learn more about range and domain of the function brainly.com/question/3412497

Answer details:

Grade: High School

Subject: Mathematics

Chapter: Polynomials

Keywords: quadratic equation, equation factorization. Factorized form, polynomial, quadratic formula, zeroes, Fundamental Theorem of algebra, polynomial, seven roots.

0 × 95 yall help if you think ppl are not smart

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Answer:0

Step-by-step explanation:

A 354-inch board is cut into two pieces. One piece is five times the length of the other. Find the length of the shorter piece

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Answer:

54 inches

Hope this helps! :)

Step-by-step explanation:

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