Class 10 Maths: Polynomials – Extra Questions with Answers

Ans: D

Solution:
For a quadratic $ax^2 + bx + c$, the sum of the roots is $-\frac{b}{a}$. In this case, $a=1$ and $b = \frac{1}{6}$. The sum of the zeros is $-\frac{1/6}{1} = -\frac{1}{6}$.
Answer: $-\frac{1}{6}$

Ans: B

Solution:
Since the roots are 2 and -4, the quadratic can be written as (x-2)(x+4). Expanding this gives x² + 4x – 2x – 8 = x² + 2x – 8.

Answer: x² + 2x – 8

Ans: B

Solution:
Let the zeroes be $\alpha, \beta, 0$.
The cubic polynomial is $2x^3 + 5x^2 + 7x = x(2x^2 + 5x + 7)$.
The product of the zeroes is $0 \cdot \alpha \cdot \beta = -\frac{0}{2} = 0$.
Since one zero is 0, the other two zeroes are the roots of $2x^2 + 5x + 7 = 0$.
The product of the roots $\alpha$ and $\beta$ of $ax^2+bx+c=0$ is $c/a$.
So the product of the other two zeroes is $\frac{7}{2}$.

Answer: 7/2

Ans: C

Solution:
$f(x) = x^2 – 4x + 3 = (x-3)(x-1)$
Setting $f(x) = 0$, we have $(x-3)(x-1) = 0$, so $x = 3$ or $x = 1$.
Answer: 1, 3

Ans: A

Solution:
Let $\alpha, \beta, \gamma$ be the zeroes of the polynomial $g(x) = 3x^3 + 2x^2 – 5ux + 7$.
The sum of the product of the zeroes taken two at a time is given by:
$\alpha\beta + \beta\gamma + \gamma\alpha = \frac{-5u}{3}$
We are given that $\alpha\beta + \beta\gamma + \gamma\alpha = -4$.
Therefore, $\frac{-5u}{3} = -4$.
Solving for $u$:
$-5u = -12$
$u = \frac{12}{5}$

Answer: $\frac{12}{5}$

Ans: B

Solution:
To find where the graph cuts the x-axis, we set $y=0$ and solve for $x$.
$0 = 2x^3 – 8x$
$0 = 2x(x^2 – 4)$
$0 = 2x(x-2)(x+2)$
Thus, $x=0, x=2, x=-2$.
Answer: (-2,0), (0,0), (2,0)

Ans: C

Solution:
Total cost of shirts = $(4x^2 + 3)(x + 2) = 4x^3 + 8x^2 + 3x + 6$
Money left = $(6x^3 + 2x^2 + 3x + 10) – (4x^3 + 8x^2 + 3x + 6) = 2x^3 – 6x^2 + 4$
If x = 1, money left = $2(1)^3 – 6(1)^2 + 4 = 2 – 6 + 4 = 0$
Answer: 0

Ans: B

Solution: Since $x^2 + 2x – 3 = (x+3)(x-1)$, we know that $x=1$ and $x=-3$ are roots.
Substituting $x=1$: $a(1)^3 + 2(1)^2 – 6(1) + b = 0 \implies a+2-6+b=0 \implies a+b=4$.
Substituting $x=-3$: $a(-3)^3 + 2(-3)^2 – 6(-3) + b = 0 \implies -27a+18+18+b=0 \implies -27a+b=-36$.
Subtracting the equations: $(a+b)-(-27a+b) = 4-(-36) \implies 28a = 40 \implies a = \frac{40}{28} = \frac{10}{7}$.
Then $b = 4 – a = 4 – \frac{10}{7} = \frac{28-10}{7} = \frac{18}{7}$.
Alternatively: Since $ax^3 + 2x^2 – 6x + b$ is divisible by $x^2 + 2x – 3$, it can be written as $(Ax+B)(x^2+2x-3)$.
$ax^3 + 2x^2 – 6x + b = Ax^3 + (2A+B)x^2 + (-3A+2B)x – 3B$.
Comparing coefficients:
$A = a$, $2A+B=2$, $-3A+2B=-6$, $-3B=b$.
$2a+B=2$, $-3a+2B=-6$. Multiply the first equation by 2: $4a+2B=4$. Subtracting the second equation: $7a = 10$, so $a=10/7$. Then $B = 2-2a = 2-20/7 = (14-20)/7 = -6/7$. Thus $b = -3B = -3(-6/7) = 18/7$.

Answer: $a = \frac{10}{7}, b = \frac{18}{7}$

Ans: B

Solution:
Let $P(x) = 3x^3 + 9ax^2 – 3bx – b – 3a + 2$. Since $(x-2)$ and $(x-1)$ are factors of $P(x)$, then $P(1)=0$ and $P(2)=0$.
$P(1) = 3(1)^3 + 9a(1)^2 – 3b(1) – b – 3a + 2 = 3 + 9a – 3b – b – 3a + 2 = 6a – 4b + 5 = 0$ (1)
$P(2) = 3(2)^3 + 9a(2)^2 – 3b(2) – b – 3a + 2 = 24 + 36a – 6b – b – 3a + 2 = 33a – 7b + 26 = 0$ (2)
From (1), $6a – 4b = -5$, so $18a – 12b = -15$ (3).
From (2), $33a – 7b = -26$. Multiply by $12/7$, $56.57a – 12b = -44.57$, which is not convenient. Multiply (2) by $4$, we get $132a – 28b = -104$. Multiply (1) by 7: $42a – 28b = -35$. Subtract to get $90a = -69$, so $a = -69/90 = -23/30$.
Substitute $a = -23/30$ into (1): $6(-23/30) – 4b + 5 = 0$, so $-23/5 – 4b + 5 = 0$. Thus $4b = 25/5 – 23/5 = 2/5$, so $b = 1/10$.
Substitute $a = -23/30$ into (2): $33(-23/30) – 7b + 26 = 0$, so $-25.3 – 7b + 26 = 0$, so $7b = 0.7$, $b = 0.1$.
From (1), $6a – 4b = -5$. From (2), $33a – 7b = -26$.
Multiply (1) by 7 and (2) by 4: $42a – 28b = -35$ and $132a – 28b = -104$. Subtract the first from the second: $90a = -69$, so $a = -69/90 = -23/30$.
Substitute $a = -23/30$ into (1): $6(-23/30) – 4b + 5 = 0$, so $-\frac{23}{5} – 4b + 5 = 0$, or $-4.6 – 4b + 5 = 0$. So $0.4 = 4b$, and $b = 1/10$.

Answer: a = -23/30, b = 1/10

Ans: B

Solution:
Since $2x^2 – x + 1$ is a factor of $ax^3 + bx – 2$, there must exist a linear factor $cx+d$ such that $(2x^2 – x + 1)(cx+d) = ax^3 + bx – 2$. Expanding the left side, we get $2cx^3 + (2d-c)x^2 + (c-d)x + d = ax^3 + bx – 2$. Comparing coefficients, we have:
$2c = a$
$2d-c = 0$
$c-d = b$
$d = -2$
From $d=-2$ and $2d-c=0$, we have $2(-2) – c = 0$, so $c = -4$.
Then $b = c – d = -4 – (-2) = -2$.
Answer: -2