Question

plz help. .........​

Answer 1

$\huge\bold\red{SOLUTION}$$\bold\purple{♡FollowMe♡}$

if x = -2 of f(x) = x³ + 13 x² + 32x + 20

(x+2) is a factor of x³ + 13x² +32x + 20

Dividing the eqns...

we got

x² + 11x + 10 = x³+13x²+32x+20 / x+2

(x² + 11x + 10)(x+2)=x² + 13x² + 32x + 20

[x² + 10x + x + 10](x+2)

[x(x+10) +1(x+10)] is a factor of p(x)

To find zeroes polyn. =0

(x+1)(x+10)=0

Either.

x= -1

or

x = -10

Hence zeroes are.

-1,-10 and -2

$<marquee>$

HOPE IT HELPS...

Answer 2

$\huge\bigstar\mathfrak\pink{\underline{\underline{SOLUTION}}}$❤️

We know that if x= a is a zero of polynomial then (x-a) is a factor of f(x).

Since -2 is zero of f(x).

Therefore, (x+2) is a factor of f(x).

Now on divide f(x)= x^3+13x^2+32x+20 by (x+2) to find other zeroes in the attachment.

By applying division algorithm, we get:

${x}^{3} + 13x {}^{2} + 32x + 20 = (x + 2)( {x}^{2} + 11x + 10)$

We do factorisation here by splitting method term,

${x}^{3} + 13 {x}^{2} + 32x + 20 = (x + 2)( {x}^{2} + 11x + 10) \\ = > {x}^{3} + {13x}^{2} + 32x + 20 = (x + 2)( {x}^{2} + 10x + x + 10) \\ = > {x}^{3} + 13 {x}^{2} + 32x + 20 = (x + 2)[x(x + 10) + 1(x + 10)] \\ = > {x}^{3} + 13 {x}^{2} + 32x + 20 = (x + 2)(x + 10)(x + 1)$

Hence, the zeroes of the given polynomial are: -2, -10 & -1

Hope it helps ☺️