Question

let p(x) = x^3 + 7x^2 + 13x + 16 , then while finding the factors of p(x) , no need to check the numbers

1) 4
2) -3
3) 2
4) 16
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Answer 1

Answer:

Hi,

Let p(x) = 2x4 - 11x³ + 7x² + 13x - 7,

It is given that ( 3 + √2) and (3-√2) are

two zeroes of p(x).

[x-(3+√2)], [x-(3-√2)] are factors

of p(x).

Now,

[x-(3+√2)][x-(3-√2)]

= [(x-3) + √2][(x-3) -√2]

=(x-3)²-(√2)²

= x² - 6x + 9-2

= x² - 6x + 7 is a factor of p(x)

x² - 6x + 7) 2x¹ -11x³ +7x² + 13x - 7(2x²+x-1 *2x² -12x³ + 14x²

*x³ - 7x² + 13x x³ -6x² + 7x

*-x² + 6x - 7

-x² + 6x - 7

By Division algorithm:

Dividend = quotient x divisor + remainder

x - 1)(x² - 6x + 7) + 0 p(x) = (2x²+x-

= [ 2x² + 2x - x - 1] (x² - 6x + 7)

= [ 2x(x + 1)-1(x+1)] (x² - 6x + 7)

= (x + 1)(2x - 1 )[x - ( 3 + √2 )][x - (3

-√2)]

Therefore,

Other two zeroes of p(x) are

x = -1, x = 1/2

I hope this helps you.

:)

Step-by-step explanation:

• please make it a brainliest answer.

Answer 2

Step-by-step explanation:

Polynomials represent the next level of algebraic complexity after quadratics. Indeed a quadratic is a polynomial of degree 2. We can factor quadratic expressions, solve quadratic equations and graph quadratic functions, the obvious question arises as to

how these things might be performed with algebraic expressions of higher degree.

The quadratic x2 − 5x + 6 factors as (x − 2)(x − 3). Hence the equation x2 − 5x + 6 = 0

has solutions x = 2 and x = 3.

Similarly we can factor the cubic x3 − 6x2 + 11x − 6 as (x − 1)(x − 2)(x − 3), which enables us to show that the solutions of x3 − 6x2 + 11x − 6 = 0 are x = 1, x = 2 or x = 3. In this module we will see how to arrive at this factorisation.

Polynomials in many respects behave like whole numbers or the integers. We can add, subtract and multiply two or