Algebraic Expressions and Identities — Exercise 10.3

Multiply:

$(5x - 2)$ by $(3x + 4)$

Answer

Solving,

$\begin{array}{rrrr} & & 5x & - 2 \\ \times & & 3x & + 4 \\ \hline & 15x^2 & - 6x & \\ & & + 20x & - 8 \\ \hline & 15x^2 & + 14x & - 8 \\ \end{array}$

Hence, the required product = $\bm{15x^2 + 14x - 8}$.

Multiply:

$(ax + b)$ by $(cx + d)$

Answer

Solving,

$\begin{array}{rrrr} & & ax & + b \\ \times & & cx & + d \\ \hline & acx^2 & + bcx & \\ & & + adx & + bd \\ \hline & acx^2 & + (ad + bc)x & + bd \\ \end{array}$

Hence, the required product = $\bm{acx^2 + (ad + bc)x + bd}$.

Multiply:

$(4p - 7)$ by $(2 - 3p)$

Answer

Solving,

Arranging the terms of the second polynomial in descending powers of $p$, we get $-3p + 2$.

$\begin{array}{rrrr} & & 4p & - 7 \\ \times & & -3p & + 2 \\ \hline & -12p^2 & + 21p & \\ & & + 8p & - 14 \\ \hline & -12p^2 & + 29p & - 14 \\ \end{array}$

Hence, the required product = $\bm{-12p^2 + 29p - 14}$.

Multiply:

$(2x^2 + 3)$ by $(3x - 5)$

Answer

Solving,

$\begin{array}{rrrrr} & & 2x^2 & & + 3 \\ \times & & & 3x & - 5 \\ \hline & 6x^3 & & + 9x & \\ & & - 10x^2 & & - 15 \\ \hline & 6x^3 & - 10x^2 & + 9x & - 15 \\ \end{array}$

Hence, the required product = $\bm{6x^3 - 10x^2 + 9x - 15}$.

Multiply:

$(1.5a - 2.5b)$ by $(1.5a + 2.5b)$

Answer

Solving,

$\begin{array}{rrrr} & & 1.5a & - 2.5b \\ \times & & 1.5a & + 2.5b \\ \hline & 2.25a^2 & - 3.75ab & \\ & & + 3.75ab & - 6.25b^2 \\ \hline & 2.25a^2 & & - 6.25b^2 \\ \end{array}$

Hence, the required product = $\bm{2.25a^2 - 6.25b^2}$.

Multiply:

$\left(\dfrac{3}{7}p^2 + 4q^2\right)$ by $7\left(p^2 - \dfrac{3}{4}q^2\right)$

Answer

Solving,

First we simplify the second polynomial as $7\Big(p^2 - \dfrac{3}{4}q^2\Big) = 7p^2 - \dfrac{21}{4}q^2$.

$\begin{array}{rrrr} & & \dfrac{3}{7}p^2 & + 4q^2 \\ \times & & 7p^2 & - \dfrac{21}{4}q^2 \\ \hline & 3p^4 & + 28p^2q^2 & \\ & & - \dfrac{9}{4}p^2q^2 & - 21q^4 \\ \hline & 3p^4 & + \dfrac{103}{4}p^2q^2 & - 21q^4 \\ \end{array}$

Hence, the required product = $\bm{3p^4 + \dfrac{103}{4}p^2q^2 - 21q^4}$.

Multiply:

$(x - 2y + 3)$ by $(x + 2y)$

Answer

Solving,

$\begin{array}{rrrrrr} & & & x & - 2y & + 3 \\ \times & & & & x & + 2y \\ \hline & x^2 & - 2xy & & + 3x & \\ & & + 2xy & - 4y^2 & & + 6y \\ \hline & x^2 & & - 4y^2 & + 3x & + 6y \\ \end{array}$

Hence, the required product = $\bm{x^2 - 4y^2 + 3x + 6y}$.

Multiply:

$(3 - 5x + 2x^2)$ by $(4x - 5)$

Answer

Solving,

Arranging the terms of the first polynomial in descending powers of $x$, we get $2x^2 - 5x + 3$.

$\begin{array}{rrrrr} & & 2x^2 & - 5x & + 3 \\ \times & & & 4x & - 5 \\ \hline & 8x^3 & - 20x^2 & + 12x & \\ & & - 10x^2 & + 25x & - 15 \\ \hline & 8x^3 & - 30x^2 & + 37x & - 15 \\ \end{array}$

Hence, the required product = $\bm{8x^3 - 30x^2 + 37x - 15}$.

Multiply:

$(3x^2 - 2x - 1)$ by $(2x^2 + x - 5)$

Answer

Solving,

$\begin{array}{rrrrrr} & & & 3x^2 & - 2x & - 1 \\ \times & & & 2x^2 & + x & - 5 \\ \hline & 6x^4 & - 4x^3 & - 2x^2 & & \\ & & + 3x^3 & - 2x^2 & - x & \\ & & & - 15x^2 & + 10x & + 5 \\ \hline & 6x^4 & - x^3 & - 19x^2 & + 9x & + 5 \\ \end{array}$

Hence, the required product = $\bm{6x^4 - x^3 - 19x^2 + 9x + 5}$.

Multiply:

$(2 - 3y - 5y^2)$ by $(2y - 1 + 3y^2)$

Answer

Solving,

Arranging the terms of both the polynomials in descending powers of $y$, we get $-5y^2 - 3y + 2$ and $3y^2 + 2y - 1$.

$\begin{array}{rrrrrr} & & & -5y^2 & - 3y & + 2 \\ \times & & & 3y^2 & + 2y & - 1 \\ \hline & -15y^4 & - 9y^3 & + 6y^2 & & \\ & & - 10y^3 & - 6y^2 & + 4y & \\ & & & + 5y^2 & + 3y & - 2 \\ \hline & -15y^4 & - 19y^3 & + 5y^2 & + 7y & - 2 \\ \end{array}$

Hence, the required product = $\bm{-15y^4 - 19y^3 + 5y^2 + 7y - 2}$.

Simplify:

$(x^2 + 3)(x - 3) + 9$

Answer

First we find the product $(x^2 + 3)(x - 3)$ by the column method, in which we write the two polynomials one below the other, multiply the first polynomial by each term of the second polynomial and then add the like terms.

$\begin{array}{rrrrr} & & x^2 & & + 3 \\ \times & & & x & - 3 \\ \hline & x^3 & & + 3x & \\ & & - 3x^2 & & - 9 \\ \hline & x^3 & - 3x^2 & + 3x & - 9 \\ \end{array}$

Now, adding 9 to the product,

$\Rightarrow x^3 - 3x^2 + 3x - 9 + 9\\[1em] \Rightarrow x^3 - 3x^2 + 3x$

Hence, $\bm{(x^2 + 3)(x - 3) + 9 = x^3 - 3x^2 + 3x}$.

Simplify:

$(x + 3) (x - 3) (x + 4) (x - 4)$

Answer

Solving,

Taking the first two factors,

$\begin{array}{rrrr} & & x & + 3 \\ \times & & x & - 3 \\ \hline & x^2 & + 3x & \\ & & - 3x & - 9 \\ \hline & x^2 & & - 9 \\ \end{array}$

Taking the last two factors,

$\begin{array}{rrrr} & & x & + 4 \\ \times & & x & - 4 \\ \hline & x^2 & + 4x & \\ & & - 4x & - 16 \\ \hline & x^2 & & - 16 \\ \end{array}$

Now, multiplying the two results,

$\begin{array}{rrrr} & & x^2 & - 9 \\ \times & & x^2 & - 16 \\ \hline & x^4 & - 9x^2 & \\ & & - 16x^2 & + 144 \\ \hline & x^4 & - 25x^2 & + 144 \\ \end{array}$

Hence, $\bm{(x + 3)(x - 3)(x + 4)(x - 4) = x^4 - 25x^2 + 144}$.

Simplify:

$(x + 5) (x + 6) (x + 7)$

Answer

Solving,

Taking the first two factors,

$\begin{array}{rrrr} & & x & + 5 \\ \times & & x & + 6 \\ \hline & x^2 & + 5x & \\ & & + 6x & + 30 \\ \hline & x^2 & + 11x & + 30 \\ \end{array}$

Now, multiplying this result by the third factor,

$\begin{array}{rrrrr} & & x^2 & + 11x & + 30 \\ \times & & & x & + 7 \\ \hline & x^3 & + 11x^2 & + 30x & \\ & & + 7x^2 & + 77x & + 210 \\ \hline & x^3 & + 18x^2 & + 107x & + 210 \\ \end{array}$

Hence, $\bm{(x + 5)(x + 6)(x + 7) = x^3 + 18x^2 + 107x + 210}$.

Simplify:

$(p + q - 2r) (2p - q + r) - 4qr$

Answer

Solving $(p + q - 2r)(2p - q + r)$,

$\begin{array}{rrrrrrr} & & & & p & + q & - 2r \\ \times & & & & 2p & - q & + r \\ \hline & 2p^2 & & & + 2pq & & - 4rp \\ & & - q^2 & & - pq & + 2qr & \\ & & & - 2r^2 & & + qr & + rp \\ \hline & 2p^2 & - q^2 & - 2r^2 & + pq & + 3qr & - 3rp \\ \end{array}$

Now, subtracting $4qr$ from the product,

$\Rightarrow (p + q - 2r)(2p - q + r) - 4qr\\[1em] \Rightarrow 2p^2 - q^2 - 2r^2 + pq + 3qr - 3rp - 4qr\\[1em] \Rightarrow 2p^2 - q^2 - 2r^2 + pq - qr - 3rp$

Hence, $\bm{(p + q - 2r)(2p - q + r) - 4qr = 2p^2 - q^2 - 2r^2 + pq - qr - 3rp}$.

Simplify:

$(p + q) (r + s) + (p - q) (r - s) - 2 (pr + qs)$

Answer

Solving,

Taking the first product,

$\begin{array}{rrrrr} & & & p & + q \\ \times & & & r & + s \\ \hline & pr & & + qr & \\ & & + ps & & + qs \\ \hline & pr & + ps & + qr & + qs \\ \end{array}$

Taking the second product,

$\begin{array}{rrrrr} & & & p & - q \\ \times & & & r & - s \\ \hline & pr & & - qr & \\ & & - ps & & + qs \\ \hline & pr & - ps & - qr & + qs \\ \end{array}$

Now, adding the two products and subtracting $2(pr + qs)$,

$\Rightarrow (pr + ps + qr + qs) + (pr - ps - qr + qs) - 2(pr + qs)\\[1em] \Rightarrow (pr + pr - 2pr) + (ps - ps) + (qr - qr) + (qs + qs - 2qs)\\[1em] \Rightarrow 0$

Hence, $\bm{(p + q)(r + s) + (p - q)(r - s) - 2(pr + qs) = 0}$.

Simplify:

$(x + y + z) (x - y + z) + (x + y - z) (-x + y + z) - 4zx$

Answer

Solving,

Taking the first product,

$\begin{array}{rrrrrrr} & & & & x & + y & + z \\ \times & & & & x & - y & + z \\ \hline & x^2 & & & + xy & & + zx \\ & & - y^2 & & - xy & - yz & \\ & & & + z^2 & & + yz & + zx \\ \hline & x^2 & - y^2 & + z^2 & & & + 2zx \\ \end{array}$

Taking the second product,

$\begin{array}{rrrrrrr} & & & & x & + y & - z \\ \times & & & & -x & + y & + z \\ \hline & - x^2 & & & - xy & & + zx \\ & & + y^2 & & + xy & - yz & \\ & & & - z^2 & & + yz & + zx \\ \hline & - x^2 & + y^2 & - z^2 & & & + 2zx \\ \end{array}$

Now, adding the two products and subtracting $4zx$,

$\Rightarrow (x^2 - y^2 + z^2 + 2zx) + (-x^2 + y^2 - z^2 + 2zx) - 4zx\\[1em] \Rightarrow (x^2 - x^2) + (-y^2 + y^2) + (z^2 - z^2) + (2zx + 2zx - 4zx)\\[1em] \Rightarrow 0$

Hence, $\bm{(x + y + z)(x - y + z) + (x + y - z)(-x + y + z) - 4zx = 0}$.

If two adjacent sides of a rectangle are $5x^2 + 25xy + 4y^2$ and $2x^2 - 2xy + 3y^2$, find its area.

Answer

Area of rectangle = length × breadth

We find the product by the column method, in which we write the two polynomials one below the other, multiply the first polynomial by each term of the second polynomial and then add the like terms.

$\begin{array}{rrrrrr} & & & 5x^2 & + 25xy & + 4y^2 \\ \times & & & 2x^2 & - 2xy & + 3y^2 \\ \hline & 10x^4 & + 50x^3y & + 8x^2y^2 & & \\ & & - 10x^3y & - 50x^2y^2 & - 8xy^3 & \\ & & & + 15x^2y^2 & + 75xy^3 & + 12y^4 \\ \hline & 10x^4 & + 40x^3y & - 27x^2y^2 & + 67xy^3 & + 12y^4 \\ \end{array}$

Hence, the area of the rectangle = $\bm{10x^4 + 40x^3y - 27x^2y^2 + 67xy^3 + 12y^4}$.