Math worksheet on polynomial functions with questions about zeros, coefficients, and graph behavior.

Since I can't view or access images directly, I’ll need you to describe the content of the image or type out the problem. However, based on your message, it appears you've provided a list of problems related to quadratic polynomials and their zeros (roots). Let me solve each of these step by step and explain the solutions.

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1. Find the zeroes of quadratic polynomial \(4x^2 - 9\) and verify the relation between the zeroes and the coefficients.

#### Step 1: Find the zeroes
We are given:
\[
4x^2 - 9 = 0
\]

This is a difference of squares:
\[
(2x)^2 - (3)^2 = 0 \Rightarrow (2x - 3)(2x + 3) = 0
\]

Set each factor equal to zero:
\[
2x - 3 = 0 \Rightarrow x = \frac{3}{2}
\]
\[
2x + 3 = 0 \Rightarrow x = -\frac{3}{2}
\]

So, the zeroes are:
\[
x = \frac{3}{2},\ -\frac{3}{2}
\]

#### Step 2: Verify the relation with coefficients

For a quadratic polynomial \(ax^2 + bx + c = 0\), the sum and product of roots are:

- Sum of roots = \(-\frac{b}{a}\)
- Product of roots = \(\frac{c}{a}\)

Here, \(a = 4\), \(b = 0\), \(c = -9\)

Sum of roots:
\[
\frac{3}{2} + \left(-\frac{3}{2}\right) = 0 = -\frac{0}{4} = 0 \quad \text{✓}
\]

Product of roots:
\[
\frac{3}{2} \times \left(-\frac{3}{2}\right) = -\frac{9}{4} = \frac{-9}{4} \quad \text{✓}
\]

✔ Verified.

---

2. Obtain the zeroes of the polynomial \(x^2 + 7x + 6\) and verify the relationship between the zeroes and the coefficients.

Factorize:
\[
x^2 + 7x + 6 = (x + 1)(x + 6) = 0
\]

So, zeroes are:
\[
x = -1,\ x = -6
\]

Now verify:

Sum of roots:
\[
-1 + (-6) = -7 = -\frac{b}{a} = -\frac{7}{1} = -7 \quad \text{✓}
\]

Product of roots:
\[
(-1)(-6) = 6 = \frac{c}{a} = \frac{6}{1} = 6 \quad \text{✓}
\]

✔ Verified.

---

3. Obtain the zeroes of \(x^2 + 2x^2 - 8x + 12\)

Wait — this seems to have a typo. The term \(x^2 + 2x^2\) suggests a cubic polynomial.

Let’s rewrite:
\[
x^2 + 2x^2 - 8x + 12 = 3x^2 - 8x + 12
\]

But the problem says “quadratic polynomial”, so likely a typo in the expression.

Possibility: It should be \(x^3 + 2x^2 - 8x + 12\)? But then it's not quadratic.

Alternatively, maybe it's supposed to be:
\[
x^2 + 2x - 8x + 12 = x^2 - 6x + 12?
\]

But that doesn’t match.

Wait — perhaps it's meant to be:
\[
x^2 + 2x - 8x + 12 = x^2 - 6x + 12
\]

But let’s assume the original is correct as written:
\[
x^2 + 2x^2 - 8x + 12 = 3x^2 - 8x + 12
\]

So we solve:
\[
3x^2 - 8x + 12 = 0
\]

Use quadratic formula:
\[
x = \frac{8 \pm \sqrt{(-8)^2 - 4(3)(12)}}{2(3)} = \frac{8 \pm \sqrt{64 - 144}}{6} = \frac{8 \pm \sqrt{-80}}{6}
\]

\[
= \frac{8 \pm 4\sqrt{-5}}{6} = \frac{4 \pm 2i\sqrt{5}}{3}
\]

So the roots are complex:
\[
x = \frac{4 + 2i\sqrt{5}}{3},\ \frac{4 - 2i\sqrt{5}}{3}
\]

But since the problem says "obtain the zeroes", and they're complex, we report them.

However, if the original was meant to be \(x^2 + 2x - 8x + 12\) → that’s still messy.

Wait — perhaps it's a typo and should be:
\[
x^2 + 2x - 8 = 0 \quad \text{or} \quad x^2 - 6x + 8 = 0?
\]

But given the current form:
\[
x^2 + 2x^2 - 8x + 12 = 3x^2 - 8x + 12
\]

We proceed with that. So no real zeroes, only complex ones.

But let's check again: perhaps the expression was meant to be:
\[
x^2 + 2x - 8 = 0? \quad \text{Then } (x+4)(x-2)=0 \Rightarrow x=-4,2
\]

But without clarification, we go with what's written.

Assuming the intended polynomial is:
\[
x^2 + 2x - 8x + 12 = x^2 - 6x + 12
\]

Still no real roots.

Alternatively, maybe it's:
\[
x^2 + 2x - 8 = 0 \quad \Rightarrow (x+4)(x-2)=0 \Rightarrow x = -4, 2
\]

But since the expression says \(x^2 + 2x^2 - 8x + 12\), which is clearly \(3x^2 - 8x + 12\), and that has no real roots, we conclude:

✔ Zeroes are complex: \(x = \frac{4 \pm 2i\sqrt{5}}{3}\)

But if the problem expects real zeroes, likely there's a typo.

Let me assume instead it was meant to be:
\[
x^2 - 2x - 8 = 0 \quad \text{(common example)}
\]

But without confirmation, I'll stick to the given.

So for \(3x^2 - 8x + 12 = 0\), the zeroes are complex.

---

4. Find a quadratic polynomial whose sum and product of zeroes are 2 and 3 respectively.

Let the polynomial be \(ax^2 + bx + c\)

We know:
- Sum of zeroes = \(-\frac{b}{a} = 2\)
- Product of zeroes = \(\frac{c}{a} = 3\)

Let \(a = 1\) (simplest case):

Then:
- \(-\frac{b}{1} = 2 \Rightarrow b = -2\)
- \(\frac{c}{1} = 3 \Rightarrow c = 3\)

So polynomial is:
\[
x^2 - 2x + 3
\]

✔ Answer: \(x^2 - 2x + 3\)

---

5. True or False

#### (a) Every quadratic polynomial has at least one zero.

✘ False

Example: \(x^2 + 1 = 0\) has no real zeroes. Even over reals, not all quadratics have real zeroes.

But over complex numbers, yes — but typically "zero" means real zero unless specified.

So in context of real numbers: False

#### (b) A cubic polynomial has 3 zeroes.

✔ True — a cubic polynomial has exactly 3 roots (real or complex, counting multiplicity), by Fundamental Theorem of Algebra.

#### (c) Degree of a polynomial is zero.

✘ False — degree of a polynomial is the highest power of \(x\). Constant non-zero polynomial has degree 0, but not all polynomials have degree zero.

The statement is ambiguous — probably false.

#### (d) Graph of \(y = 4x + 6\) meets x-axis at 2 points.

✘ False — this is a linear equation. It intersects x-axis at one point only.

Solve: \(4x + 6 = 0 \Rightarrow x = -\frac{3}{2}\)

Only one zero.

#### (e) Sum of zeroes of a quadratic polynomial always has two distinct zeroes.

✘ False — it can have repeated roots.

Example: \(x^2 - 2x + 1 = (x-1)^2\) has double root at \(x=1\)

So not necessarily two distinct zeroes.

#### (f) A quadratic may meet x-axis only at one point.

✔ True — when discriminant is zero, it touches x-axis at one point (repeated root).

#### (g) If the zeroes of a quadratic polynomial \(ax^2 + bx + c\) are both negative, then \(a, b, c\) have the same sign.

Let’s test.

Suppose both roots are negative: say \(-p, -q\), \(p,q > 0\)

Then:
- Sum = \(-p - q = -(p+q)\) < 0 ⇒ \(-\frac{b}{a} < 0\) ⇒ \(\frac{b}{a} > 0\) ⇒ \(a\) and \(b\) same sign
- Product = \((-p)(-q) = pq > 0\) ⇒ \(\frac{c}{a} > 0\) ⇒ \(a\) and \(c\) same sign

So \(a, b, c\) all have same sign if \(a > 0\): then \(b > 0\), \(c > 0\)

If \(a < 0\), then \(b < 0\), \(c < 0\)

So yes — all three have same sign.

✔ True

#### (h) Zeroes of polynomial \(x^2 - 4\) are equal in magnitude but opposite in sign.

Yes: \(x^2 - 4 = (x-2)(x+2)\), roots: \(2, -2\)

Magnitude: both 2, signs opposite.

✔ True

#### (i) Polynomial \(x^2 + 1\) has only two real zeroes.

No — \(x^2 + 1 = 0 \Rightarrow x = \pm i\), no real zeroes.

So it has zero real zeroes, not two.

✘ False

#### (j) Sum of zeroes in polynomial \(x^2 + ax + a\) is same as the product of its zeroes.

Let’s compute:

Sum of zeroes = \(-\frac{a}{1} = -a\)

Product = \(\frac{a}{1} = a\)

Are they equal?

\[
-a = a \Rightarrow 2a = 0 \Rightarrow a = 0
\]

Only true when \(a = 0\), not always.

So not always true.

✘ False

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✔ Final Answers Summary:

1. Zeroes of \(4x^2 - 9\): \(\boxed{\frac{3}{2}},\ \boxed{-\frac{3}{2}}\); verified.
2. Zeroes of \(x^2 + 7x + 6\): \(\boxed{-1},\ \boxed{-6}\); verified.
3. Zeroes of \(3x^2 - 8x + 12\): \(\boxed{\frac{4 \pm 2i\sqrt{5}}{3}}\) (complex).
4. Quadratic with sum 2, product 3: \(\boxed{x^2 - 2x + 3}\)
5. True/False:
- (a) False
- (b) True
- (c) False
- (d) False
- (e) False
- (f) True
- (g) True
- (h) True
- (i) False
- (j) False

Let me know if you want me to revise any part based on corrected expressions!