Question

FACTORISE 9X3-3X2-5X-1

Answer 1

Answer:

Step-by-step explanation:

(X-1 )(3x+1)(3x+1) are factors of this equation

Answer 2

Factors of $\bf 9 {x}^{3} - 3 {x}^{2} - 5x - 1$ are $\bf \red{(x - 1)( {3x + 1)}^{2}}$.

Given:

• $9 {x}^{3} - 3 {x}^{2} - 5x - 1$

To find:

• Factors of cubic polynomial.

Solution:

Concept/Theorem to be used:

• Find first factor by factor theorem(hit and trial method)
• Factor theorem: It states that if x-a is a factor of polynomial p(x), then p(a)=0.
• Divide the polynomial by factor.
• Factorise the quotient polynomial.

Step 1:

Apply factor theorem.

Let the polynomial is p(x).

$p(x) = 9 {x}^{3} - 3 {x}^{2} - 5x - 1$

put

x=1

$p(1) = 9 {(1)}^{3} - 3 {(1)}^{2} - 5(1) - 1$

or

$= 9- 3 - 5 - 1$

or

$\bf p(1) = 0$

Thus,

x=1 is a zero of polynomial.

So, (x-1) is a factor.

Step 2:

Divide p(x) by x-1.

$\quad \quad x - 1 \: ) \: 9 {x}^{3} - 3 {x}^{2} - 5x - 1 \: (9 {x}^{2} + 6x + 1\\ 9 {x}^{3} - 9 {x}^{2} \quad \: \: \: \: \: \: \: \: \: \: \: \: \\ ( - ) \: \: \: \: \: ( + ) \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ - - - - - - - \: \: \: \: \: \: \: \\ 6 {x}^{2} - 5x \\ 6 {x}^{2} - 6x \\ ( - ) \: \: ( + ) \\ - - - - - \\ x - 1 \\ x - 1 \\ ( - ) \: \: ( + ) \\ - - - - - \\ 0 \\ - - - - -$

Thus,

Quotient polynomial is $\bf 9 {x}^{2} + 6x + 1$

Step 3:

Factorise the quotient polynomial.

$= 9 {x}^{2} + 3x + 3x + 1$

or

$= 3x(3x + 1) + 1(3x + 1)$

or

$9 {x}^{2} + 6x + 1= (3x + 1)(3x + 1)$

Thus,

Factors of given polynomial are (x-1)(3x+1)².

Learn more:

1) X3-7x-6. Factorise the polynomial

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2) find the zeros of the cubic polynomial 3 x cube minus 5 x square - 11 x minus 3

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