Important Questions (1 Mark): Polynomials - Grade 10 MCQ

Sum of zeros = 3/1
-b/a = 3/1
Product of zeros = 2/1
c/a = 2/1
This gives 
a = 1
b = -3
c = -2,
The required quadratic equation is
ax²+bx+c
So,  x²-3x+2

Given ,g(x)=x+2=0
x=-2

P(x)=x³-2ax²+16
p(-2)=(-2³)-2×a×(-2²)+16
= -8 - 8a +16
= 8 - 8a
⇒ -8a = -8
⇒ a=1
Hence, value of a=1

So option B is correct answer. 

Given :
3 + 5 + (-8) = 0.

Then,
3³ + 5³ + (-8)³ = ?

Using, identity...
If a+b+c=0, then a³ + b ³ + c³ = 3abc

Let a = 3, b = 5 and c = -8.

Here, a + b + c = 3 + 5 + (-8) = 0.

Then, a³ + b³ + c³ = 3³ + 5³ + (-8)³ 
= 3× 3 × 5 × -8
= 9 × -40
= -360

Using mid-term splitting,
x²+x-20=x²+5x-4x-20=x(x+5)-4(x+5)
Taking common x+5
(x+5)(x-4) , so the other factor is x-4

P(x) = 2x² + 5x + 1
Sum of roots = -5/2
Product of roots = 1/2
Therefore substituting these values, 
α + β +αβ 
=(α + β) + αβ
= -5/2 + 1/2
= -4/2 
= -2

If α  and β are the zeros of the polynomial then
(x−α)(x−β) are the factors of the polynomial
Thus, (x−α)(x−β) is the polynomial.
So, the polynomial =x² − αx − βx + αβ
=x² − (α + β)x + αβ....(i) 
Now,the quadratic polynomial is  
2 − 3x − x² = x² + 3x − 2....(ii)

Now, comparing equation (i) and (ii),we get,
−(α + β) = 3 
α + β = −3

The quadratic equation is of the form  x² - (sum of zeros) x + (product of zeros)
=x² - 5x - 14

Zeros of the polynomials are the values which gives zero when their value is substituted in the polynomial
When x=2,
x²+ax+b =(2)²+a*2+b=0
4+2a+b=0
b=-4-2a    ….1
When x=3,
(3)²+ 3a + b=0
9 + 3a + b=0
Substituting 
9 + 3a - 4 - 2a =0
5 + a =0
a = -5
b = 6

We have remainder theorem as 
x³-3x²+x+2=q(x)*g(x)+r(x)
x³-3x²+x+2=(x-2)g(x)+(4-2x)
g(x)= 
So by division method,
g(x)= x² - x + 1

The correct option is B 1
Let p(x)=x³−2ax²+16
and g(x)=x+2.
If x - a is a factor of p(x), then p(a)=0.
Put g(x)=0
x+2=0
x=−2
p(−2)=(−2)³−2a(−2)²+16=0
⇒−8−8a+16=0
⇒a=1

We are given the cubic polynomial:

f(x) = 3x³ - 5x² - 11x - 3

It is stated that one root of the polynomial is:

x = -1/3

Step 1: Divide the polynomial by (3x + 1)

Since x = -1/3 is a root, (3x + 1) is a factor of f(x).

Using polynomial long division:

f(x) = (3x + 1)(x² - 2x - 3)

Step 2: Factorize the quadratic

The quadratic factor is:

x² - 2x - 3

Factorizing it:

x² - 2x - 3 = (x - 3)(x + 1)

Step 3: Identify the roots

The roots of the polynomial are:

• x = -1/3
• x = 3
• x = -1

Final Answer

The other two roots are:

3 and -1

Correct option: (c) 3, -1

Given α and β are the zeroes of the polynomial x² − 5x + k
Also given that α − β = 1 → (1)
Recall that sum of roots (α + β) = −(b/a)
∴ α + β = 5 → (2)
Add (1) and (2), we get
α − β = 1
α + β = 5
2α = 6
∴ α = 3
Put α = 3 in α + β = 5
3 + β = 5
∴ β = 2
Hence 3 and 2 are zeroes of the given polynomial
Put x = 2 in the given polynomial to find the value of k ( Since 2 is a zero of the polynomial, f(2) will be 0 )

x² − 5x + k = 0
⇒ 2² − 5(2) + k = 0
⇒ 4 − 10 + k = 0
⇒ − 6 + k = 0
∴ k = 6

So it s an equation of a parabola.

The graph will be that of a parabola.

The correct answer is a.

(1+1/α) (1+1/β) 

= (α+1/α) (β+1/β)

=αβ + α + β + 1 / αβ

=( 2/5 + 1/3+ 1)  /  2/5

=26/15 X 5/2 = 13/3

Sum of two solutions is zero ,so both are inverse of each other. So we have the solutions as a,-a,b.
(x-a)(x+a)(x+b)=(x²-a²)(x+b)=x³+bx²-a2x-a2b
pq=-a²b=r

Let P(x)=(x−a)(x+a)Q(x)+rx+s
When we divide by x+a, we get
P(−a)=(−a−a)(−a+a)Q(−a)+r(−a)+s=a
⇒s−ra=a   ...(1)
And when we divide by x−a, we get
P(a)=(a−a)(a+a)Q(a)+ra+s=−a
⇒s+ra=−a    ...(2)
Solving (1) and (2), we get
s=0,r=−1
Hence, P(x)=(x−a)(x+a)Q(x)−x
And when we divide by x2-a2, it leaves a remainder −x.

Let one root be a, other root is a².
(x-a)(x-a²)=x²+(-a²-a)x+a³
p³=(-a²-a)³=-a⁶-3a⁵-3a⁴-a³
-q(3p-1)= -a³(3(-a²-a)-1)=3a⁵+3a⁴+a³
q²=a⁶
Substituting the values
-a⁶-3a⁵-3a⁴-a³+3a⁵+3a⁴+a³+a⁶=0

Alpha(a) and beta(b) are roots of x^2 + px + 1

This implies that sum of roots= a+b = -p/1=-p

And the product of roots = ab = 1/1=1

Similarly ,

Gamma(c) and delta(d) are roots of x^2 + qx + 1

So c+d=-q and cd =1.

The above results can be obtained once we know that any quadratic equation has two roots and hence can be written as (x-p)(x-q)=0 where a and b are the roots .

So x^2 -(p+q)x + pq =0

Comparing this with the general form of quadratic equation :ax^2 + bx + c= 0 we get

Sum of the roots =p+q= -b/a

And product of the roots = pq = c/a}
 

RHS=(a-c)(b-c)(a+d)(b+d)

=(c^2-ac-bc +ab)(d^2 +bd +ad + ab)

We know ab=1

So RHS= (c^2-ac-bc +1)(d^2 +bd +ad + 1)

= (c^2)(d^2) +(a+b)c^2(d) + c^2 -(d^2)c(a+b) -(a+b)^2(cd) -(a+b)c + d^2 + bd + ad + 1

= 1 + ac+bc + c^2 - da-db - (a^2 + b^2 + 2(1)) -ac -bc + bd + ad +1

Cancelling off all the common terms,

We get c^2 +d^2-a^2-b^2

= c^2+d^2-a^2-b^2
=c^2 -a^2 + d^2 -b^2 + 2(1) -2(1){ ab=cd =1}
=c^2 + d^2 + 2cd - a^2 - b^2 - 2ab
LHS=(c+d)^2-(a+b)^2
Therefore,
But we know -p= a+b and -q = c+d
LHS= q^2-p^2

Let α and β be the roots of the given equation.
Then, α + β = 5/2 and αβ = 2/2 = 1
∴ 2α + 2β
∴ 2(α + β)
∴ 5,(2α)(2β) = 4
So, the requiared equation is : 
x²−5x+4=0