Class 9th math 4.4 solution english PCTB

Note: This is the Solution of review exercise 4.4 from newly published book by PCTB (Punjab Curriculum and Textbook Board, Pakistan) for new 9th session 2025 Onward.

Exercise 4.4

Question # 01: Find the square root of the following polynomials by factorization method:

\((i)\) \(\quad\) \( x^2-8x+16\)

\((ii)\) \(\quad\) \(9x^2+12x+4\)

\((iii)\) \(\quad\) \(36a^2+84a+49 \)

\((iv)\) \(\quad \)\(64y^2-32y+4 \)

\((v)\) \(\quad \) \(200t^2-120t+18\)

\((vi)\) \(\quad \) \(40x^2+120x+90\)

Question 2: Find the square root of the following polynomials by division method:

\((i)\) \(\quad\) \( 4x^4-28x^3+37x^2+42x+9\)

\((ii)\) \(\quad\) \(121x^4-198x^3-183x^2+216x+144\)

\((iii)\) \(\quad\) \(x^4-10x^3y+27x^2y^2-10xy^3+y^4 \)

\((iv)\) \(\quad \)\(4x^4-12x^3+37x^2-42x+49 \)

Question 3: An investor’s return \(\small{R(x)}\) in rupees after investing \(\small{x}\) thousand rupees is given by quadratic expression:

\[R(x)=-x^2+6x-8\]

Factorize the expression and find the investment levels that result in zero return.

Question 4: A company’s profit \(\small{P(x)}\) in rupees from selling \(\small{x}\) units of a product is modeled by the cubic expression:

\[P(x)=-x^3-15x^2+75x-125\]

Find the break-even point(s), where the profit is zero.

Question 5: The potential energy \(\small{V(x)}\) in an electric field varies as cubic function of distance \(\small{x}\), given by:

\[V(x)=2x^3-6x^2+4x\]

Determine where the potential energy is zero.

Question 6: In structural engineering, the deflection \(\small{Y(x)}\) of a beam is given by:

\[Y(x)=2x^2-8x+6\]

This equation gives the vertical deflecyion at any point \(\small{x}\) along the beam. Find the points of zero deflection.

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