Question

Find the integral roots of the polynomial ${f(x)} ={x}^{3} + {6x}^{2} + {11x + 6}$.​

Answer 1

Answer:

$\</strong><strong>small</strong><strong>\</strong><strong>bold</strong><strong>{\pink{\green {\sf\boxed{</strong><strong> Integral </strong><strong>\</strong><strong>:</strong><strong>roots</strong><strong>\</strong><strong>:</strong><strong> </strong><strong>of</strong><strong>\</strong><strong>:</strong><strong> </strong><strong>fx</strong><strong>\</strong><strong>:</strong><strong> </strong><strong>are</strong><strong>\</strong><strong>:</strong><strong> -1, -2, -3.</strong><strong>}</strong><strong>}}}$

Step-by-step explanation:

$\bold{\underline {Given:}}$

${f(x)} ={x}^{3} + {6x}^{2} + {11x + 6}$

$\bold{\underline {To \:Find:}}$

❋ Integral roots of the given polynomial =?

$\huge {\red{\boxed{SOLUTION⟹}}}$

${f(x)} ={x}^{3} + {6x}^{2} + {11x + 6}$

Clearly ${fx}$ is polynomial with integral coefficients and the coefficient of the highest degree term, i. e, the leading Coefficient is one there for integral roots of ${fx}$ are limited to the integral factors of 6, which are + - 1, + - 2, + - 3, + - 6. [ Note this step]

We observe that :

${f(-1)}$ = - 1 + 6 - 11 + 6 = 0

${f(-2)}$ = - 8 + 24 - 22 + 6 = 0

${f(-3)}$ = - 27 + 54 - 33 + 6 = 0

$\Large\boxed{Hence}$

•°• Integral roots of ${fx}$ are -1, -2, -3.

#Be_Brainly

Answer 2

Answer:

Let p(x) = x 3 + 6x 2 + 11x + 6

Put x = – 1

p(– 1) = (– 1)3 + 6(– 1)2 + 11(– 1) + 6 = – 1 + 6 – 11 + 6 = 0

∴ (x + 1) is a factor of p(x)

So, we cam break up terms of p(x) as follows.

p(x) = x 3 + 6x 2 + 11x + 6



Therefore, the integral roots of the given equation is find out as:

p(x) = 0

⇒ (x+1)(x+2)(x+3) = 0

⇒ x = -1, -2, -3