Math, asked by aj087981, 11 months ago

Using factor theorem,factorize 2x^3 -3x^2 -7x+30.

Answers

Answered by vishal5323
8

Answer:

2x3+x2-61x+30=0

Three solutions were found :

x = 5

x = -6

x = 1/2 = 0.500

Step by step solution :

Step 1 :

Equation at the end of step 1 :

((2x3 + x2) - 61x) + 30 = 0

Step 2 :

Checking for a perfect cube :

2.1 2x3+x2-61x+30 is not a perfect cube

Trying to factor by pulling out :

2.2 Factoring: 2x3+x2-61x+30

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

Group 1: -61x+30

Group 2: 2x3+x2

Pull out from each group separately :

Group 1: (-61x+30) • (1) = (61x-30) • (-1)

Group 2: (2x+1) • (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 :

2.3 Find roots (zeroes) of : F(x) = 2x3+x2-61x+30

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 2 and the Trailing Constant is 30.

The factor(s) are:

of the Leading Coefficient : 1,2

of the Trailing Constant : 1 ,2 ,3 ,5 ,6 ,10 ,15 ,30

Let us test ....

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

-1 1 -1.00 90.00

-1 2 -0.50 60.50

-2 1 -2.00 140.00

-3 1 -3.00 168.00

-3 2 -1.50 117.00

-5 1 -5.00 110.00

-5 2 -2.50 157.50

-6 1 -6.00 0.00 x+6

-10 1 -10.00 -1260.00

-15 1 -15.00 -5580.00

-15 2 -7.50 -300.00

-30 1 -30.00 -51240.00

1 1 1.00 -28.00

1 2 0.50 0.00 2x-1

2 1 2.00 -72.00

3 1 3.00 -90.00

3 2 1.50 -52.50

5 1 5.00 0.00 x-5

5 2 2.50 -85.00

6 1 6.00 132.00

10 1 10.00 1520.00

15 1 15.00 6090.00

15 2 7.50 472.50

30 1 30.00 53100.00

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

2x3+x2-61x+30

can be divided by 3 different polynomials,including by x-5

Polynomial Long Division :

2.4 Polynomial Long Division

Dividing : 2x3+x2-61x+30

("Dividend")

By : x-5 ("Divisor")

dividend 2x3 + x2 - 61x + 30

- divisor * 2x2 2x3 - 10x2

remainder 11x2 - 61x + 30

- divisor * 11x1 11x2 - 55x

remainder - 6x + 30

- divisor * -6x0 - 6x + 30

remainder 0

Quotient : 2x2+11x-6 Remainder: 0

Trying to factor by splitting the middle term

2.5 Factoring 2x2+11x-6

The first term is, 2x2 its coefficient is 2 .

The middle term is, +11x its coefficient is 11 .

The last term, "the constant", is -6

Step-1 : Multiply the coefficient of the first term by the constant 2 • -6 = -12

Step-2 : Find two factors of -12 whose sum equals the coefficient of the middle term, which is 11 .

-12 + 1 = -11

-6 + 2 = -4

-4 + 3 = -1

-3 + 4 = 1

-2 + 6 = 4

-1 + 12 = 11 That's it

Step-3 : Rewrite the polynomial splitting the middle term using the two factors found in step 2 above, -1 and 12

2x2 - 1x + 12x - 6

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

x • (2x-1)

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

6 • (2x-1)

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

(x+6) • (2x-1)

Which is the desired factorization

Equation at the end of step 2 :

(2x - 1) • (x + 6) • (x - 5) = 0

Step 3 :

Theory - Roots of a product :

3.1 A product of several terms equals zero.

When a product of two or more terms equals zero, then at least one of the terms must be zero.

We shall now solve each term = 0 separately

In other words, we are going to solve as many equations as there are terms in the product

Any solution of term = 0 solves product = 0 as well.

Answered by anoopsingh3467
0

Answer:

Factorize: 2x3 – 3x2

-17x +30

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