Question

X3-7x-6. Factorise the polynomial

Answer 1

HOPE IT HELPS .....

THANKS.....

BY $$ ALIA.....

Answer 2

The factors of $\bf {x}^{3} - 7x - 6$

are $\bf (x + 1),(x + 2)\: and\:(x - 3)$

Given:

• ${x}^{3} - 7x - 6$

To find:

• Factors of cubic polynomial.

Solution:

Concept/Theorem to be used:

• Factor theorem:If x-a is a factor of p(x), then p(a)=0.
• Using factor theorem find one of the factor, then divide the polynomial by that factor and then factorise the quotient polynomial.

Step 1:

Find the factor of polynomial using hit and trial method.

Let

$p(x) = {x}^{3} - 7x - 6$

put x= -1.

$p( - 1) = {( - 1)}^{3} - 7( - 1) - 6$

or

$p( - 1) = - 1 + 7 - 6$

or

$p( - 1) = 7 - 7$

or

$\bf p( - 1) = 0$

Thus,

(x+1) is a factor of p(x).

Step 2:

Divide p(x) by (x+1).

$x + 1 \: ) \: {x}^{3} - 7x - 6 \: ( \: {x}^{2} - x - 6 \\ \: \: \: \: \: \: \: \: \: {x}^{3} \: \: \: \: \: \: \: \: \: \: \: \: + {x}^{2} \\ ( - ) \: \: \: \: \: \: \: \: ( - ) \\ - - - - - - - - - \\ - {x}^{2} - 7x - 6 \\ - {x}^{2} - x \: \: \: \: \: \: \: \: \\ ( + ) \: \: \: ( + ) \: \: \: \: \: \: \: \: \: \\ - - - - - - - \\ - 6x - 6 \\ - 6x - 6 \\ ( + ) \: \: ( + ) \\ - - - - - - - \\ 0 \\ - - - - - - -$

Quotient polynomial is $\bf {x}^{2} - x - 6$

Step 3:

Factorise the quotient polynomial.

Split the middle term.

$= {x}^{2} - 3x + 2x - 6$

or

$= {x}(x - 3) + 2(x - 3)$

or

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

Thus,

All three factors of given cubic polynomial are (x+1), (x+2) and (x-3).

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