Question

if X + 5 is a factor of x cube + 2 x square - 13 x + 10 then find the other factors​

Answer 1

Answer:

step 1 __If x+5 is a factor...

p(x)= x³+2x²-13x+10

p(-5) = -5³+2(-5²)-13*-5+10

0= -125 +2*25 +65 +10

0= -125 +50+65+10

0= -125 +125

0=0

yes.. one factor is (x+5)

step 2 ___ by long division.(see long division in the attachment).

after doing long division the quotient will be ...

x²-3x+2

step 3___ by middle term splitting

x²-3x+2

x²-x-2x+2

x(x-1) -2(x-1)

(x-1)(x-2)

so..the factors are

(x+5)(x-1)(x-2)

thank you

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Answer 2

Question: If $x+5$ is a factor of $x^3+2x^2-13x+10$ then find the other factors.

Answer:

$x-2$ and $x-1$ are the other two factors of the polynomial $x^3+2x^2-13x+10$.

Step-by-step explanation:

Consider the polynomial as follows:

$x^3+2x^2-13x+10$ . . . . . (1)

Adding and subtracting the terms $3x^2$ and $2x$ in the polynomial (1) as follows:

⇒ $x^3+2x^2 +(3x^2-3x^2)-13x+(-2x+2x)+10$

⇒ $x^3+(2x^2 +3x^2)-3x^2+(-13x-2x)+2x+10$

Rewrite as follows:

⇒ $x^3+5x^2 -3x^2-13x+2x+10$

⇒ $(x^3+5x^2) -(3x^2+13x)+(2x+10)$

Further, simplify as follows:

⇒ $x^2(x+5) -3x(x+5)+2(x+5)$

⇒ $(x+5)(x^2-3x+2)$

This implies $(x+5)$ is factor one of the factor of $x^3+2x^2-13x+10$ which is already given in the question.

Now,

Using middle term splitting method,

Factorise the quadratic polynomial $(x^2-3x+2)$ as follows:

⇒ $x^2-2x-x+2$

⇒ $x(x-2)-1(x-2)$

⇒ $(x-2)(x-1)$

Therefore, the other two factors are $x-2$ and $x-1$.

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