Question

here you need to factorise the polynomials using factor theroem
pls...i need the answer
unwanted answer would be reported if possible pls give the answer with an attachment...pls​

Answer 1

Answer:

1). Let,

f(x)= x⁴+10x³+35x²+50x+24

Now putting x=1, we get

f(-1)= (-1)⁴+10(-1)³+35(-1)+24= 1-10+35-50+24= 60-60

=0

Therefore,(x) is a factor of polynomial

f(x)= x³(x+1)9x²(x+1)+26x(x+1)

= (x+1) {x³+9x²+26x+24}

= (x+1)g(x)..1eq

where g(x)=x³+9x²+26x+24

putting x= -2 we get

g(-2)=(-2)³+9(-2)²+26(-2)+2

= -8 +36-52+24= 60-60

=0

Therefore, (x+2) is the factor of g(x)

Now,

g(x)=x²(x+2)+7x(x+2)+12(x+2)

(x=2) {x²+4x+3x+12)}

(x+2(x+3(x+4)..2 eq

from eq1&2 we get

f(x)= (x+1)(x+2)(x+3)(x+4)

hence (x+1), (x+2), (x+3)&(x+4) are the factors of polynomial f(x)

2). x=1 ,2,-2,3.

Step-by-step explanation:

How do I factorize x4−2x3−7x2+8x+12 ?

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By trial and error the roots are factors of 12 that is -12,-6,-4,-3,-2,-1,0,1,2,3,4,6,12.

Let f(x)=x⁴-2x³-7x²+8x+12

f(-1)=(-1)⁴-2(-1)³-7(-1)²+8(-1)+12=0

So (x+1) is a factor.

Using synthetic division

-1| 1 -2 -7 8 12

| 0 -1 3 4 -12

|1. -3 -4 12 0*

Where 0* is a zero only but it indicates the remainder of the division.

So the polynomial is written as

x⁴-2x³-7x²+8x+12=(x+1)(x³-3x²-4x+12)

Now consider,

g(x)=x³-3x²-4x+12

similarly, if we test as above

g(-2)=(-2)³-3(-2)²-4(-2)+12=-8–12+8+12=0

Now again computing synthetic division

-2| 1 -3 -4. 12

|0 -2 10 -12

|1 -5 6 0

So the g(x)=(x+2)(x²-5x+6)

x²-5x+6=x²-2x-3x+6=x(x-2)-3(x-2)=(x-2)(x-3)

g(x)=(x+2)(x-2)(x-3)

f(x)=x⁴-2x³-7x²+8x+12

f(x)=(x+1)g(x)

f(x)=(x+1)(x+2)(x-2)(x-3).

3). 45⇒±1,±3,±5,±9,±15,±45

If we put x = 1 in p(x)

p(1)=2(1)4−7(1)3−13(1)2+63(1)−45

p(1)=2−7−13+63−45=65−65=0

∴x=1 or x - 1 is a factor of p(x).

Similarly if we put x = 3 in p(x)

p(3)=2(3)4−7(3)3−13(3)2+63(3)−45

p(3)=162−189−117+189−45=162−162=0

Hence, x = 3 or (x - 3) = 0 is the factor of p(x).

p(x)=2x4−7x3−13x2+63x−45

∴p(x)=2x3(x−1)−5x2(x−1)−18x(x−1)+45(x−1)

⇒p(x)=(x−1)(2x3−5x2−18x+45)

⇒p(x)=(x−1)(2x3−5x2−18x+45)

⇒p(x)=(x−1)[2x2(x−3)+x(x−3)−15(x−3)]

⇒p(x)=(x−1)(x−3)(2x2+x−15)

⇒p(x)=(x−1)(x−3)(2x2+6x−5x−15)

⇒p(x)=(x−1)(x−3)[2x(x+3)−5(x+3)]

⇒p(x)=(x−1)(x−3)(x+3)(2x−5).

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