Question

(h) $9 - x^{2}$
$(3+x)(3-x)$
(i) $4x^{2} - 1$
$(2x+1)(2x-1)$
(j) $16x^{2} - 49$
$(4x-7)(4x+7)$
(k) $1 - 25x^{2}$
$(1+5x)(1-5x)$
(l) $x^{2} - 9y^{2}$
$(x-3y)(x+3y)$

Explanation:

Step1: Recognize difference of squares

Recall the identity: $a^2 - b^2 = (a+b)(a-b)$

Step2: Factor (f) $9 - x^2$

Rewrite as $3^2 - x^2$, so:
$9 - x^2 = (3+x)(3-x)$

Step3: Factor (g) $4x^2 - 1$

Rewrite as $(2x)^2 - 1^2$, so:
$4x^2 - 1 = (2x+1)(2x-1)$

Step4: Factor (h) $16x^2 - 49$

Rewrite as $(4x)^2 - 7^2$, so:
$16x^2 - 49 = (4x+7)(4x-7)$

Step5: Factor (i) $1 - 25x^2$

Rewrite as $1^2 - (5x)^2$, so:
$1 - 25x^2 = (1+5x)(1-5x)$

Step6: Factor (j) $x^2 - 9y^2$

Rewrite as $x^2 - (3y)^2$, so:
$x^2 - 9y^2 = (x+3y)(x-3y)$

Answer:

(f) $\boldsymbol{(3+x)(3-x)}$
(g) $\boldsymbol{(2x+1)(2x-1)}$
(h) $\boldsymbol{(4x+7)(4x-7)}$
(i) $\boldsymbol{(1+5x)(1-5x)}$
(j) $\boldsymbol{(x+3y)(x-3y)}$