English, asked by behapppyyy, 9 months ago

factorise x^3+5x^2+3x-9​

Answers

Answered by guptapreeti051181
7

Answer:

Step by Step Solution:

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STEP

1

:

Equation at the end of step 1

(((0 - (x3)) + 5x2) - 3x) - 9

STEP

2

:

STEP

3

:

Pulling out like terms

3.1 Pull out like factors :

-x3 + 5x2 - 3x - 9 =

-1 • (x3 - 5x2 + 3x + 9)

Checking for a perfect cube :

3.2 x3 - 5x2 + 3x + 9 is not a perfect cube

Trying to factor by pulling out :

3.3 Factoring: x3 - 5x2 + 3x + 9

Thoughtfully split the expression at hand into groups, each group having two terms :

Group 1: 3x + 9

Group 2: x3 - 5x2

Pull out from each group separately :

Group 1: (x + 3) • (3)

Group 2: (x - 5) • (x2)

Bad news !! Factoring by pulling out fails :

The groups have no common factor and can not be added up to form a multiplication.

Polynomial Roots Calculator :

3.4 Find roots (zeroes) of : F(x) = x3 - 5x2 + 3x + 9

Polynomial Roots Calculator is a set of methods aimed at finding values of x for which F(x)=0

Rational Roots Test is one of the above mentioned tools. It would only find Rational Roots that is numbers x which can be expressed as the quotient of two integers

The Rational Root Theorem states that if a polynomial zeroes for a rational number P/Q then P is a factor of the Trailing Constant and Q is a factor of the Leading Coefficient

In this case, the Leading Coefficient is 1 and the Trailing Constant is 9.

The factor(s) are:

of the Leading Coefficient : 1

of the Trailing Constant : 1 ,3 ,9

Let us test ....

P Q P/Q F(P/Q) Divisor

-1 1 -1.00 0.00 x + 1

-3 1 -3.00 -72.00

-9 1 -9.00 -1152.00

1 1 1.00 8.00

3 1 3.00 0.00 x - 3

9 1 9.00 360.00

The Factor Theorem states that if P/Q is root of a polynomial then this polynomial can be divided by q*x-p Note that q and p originate from P/Q reduced to its lowest terms

In our case this means that

x3 - 5x2 + 3x + 9

can be divided by 2 different polynomials,including by x - 3

Polynomial Long Division :

3.5 Polynomial Long Division

Dividing : x3 - 5x2 + 3x + 9

("Dividend")

By : x - 3 ("Divisor")

dividend x3 - 5x2 + 3x + 9

- divisor * x2 x3 - 3x2

remainder - 2x2 + 3x + 9

- divisor * -2x1 - 2x2 + 6x

remainder - 3x + 9

- divisor * -3x0 - 3x + 9

remainder 0

Quotient : x2-2x-3 Remainder: 0

Trying to factor by splitting the middle term

3.6 Factoring x2-2x-3

The first term is, x2 its coefficient is 1 .

The middle term is, -2x its coefficient is -2 .

The last term, "the constant", is -3

Step-1 : Multiply the coefficient of the first term by the constant 1 • -3 = -3

Step-2 : Find two factors of -3 whose sum equals the coefficient of the middle term, which is -2 .

-3 + 1 = -2 That's it

Step-3 : Rewrite the polynomial splitting the middle term using the two factors found in step 2 above, -3 and 1

x2 - 3x + 1x - 3

Step-4 : Add up the first 2 terms, pulling out like factors :

x • (x-3)

Add up the last 2 terms, pulling out common factors :

1 • (x-3)

Step-5 : Add up the four terms of step 4 :

(x+1) • (x-3)

Which is the desired factorization

Multiplying Exponential Expressions:

3.7 Multiply (x-3) by (x-3)

The rule says : To multiply exponential expressions which have the same base, add up their exponents.

In our case, the common base is (x-3) and the exponents are :

1 , as (x-3) is the same number as (x-3)1

and 1 , as (x-3) is the same number as (x-3)1

The product is therefore, (x-3)(1+1) = (x-3)2

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