Using (x + y + z)² = x² + y² + z² + 2xy + 2yz + 2zx, expand (a + 2b + 3c)².

a² + 4b² + 9c² + 4ab + 12bc + 6ca. Squares: a², (2b)² = 4b², (3c)² = 9c². Cross terms, each doubled: 2(a)(2b) = 4ab, 2(2b)(3c) = 12bc, 2(3c)(a) = 6ca. So (a + 2b + 3c)² = a² + 4b² + 9c² + 4ab + 12bc + 6ca.

Using $(x + y + z)^{2} = x^{2} + y^{2} + z^{2} + 2 x y + 2 yz + 2 z x$ , expand $(a + 2 b + 3 c)^{2}$ .

Class 9PolynomialsPolynomials

A $a^{2} + 4 b^{2} + 9 c^{2} + 4 ab + 12 b c + 6 c a$
B $a^{2} + 4 b^{2} + 9 c^{2} + 2 ab + 6 b c + 3 c a$
C $a^{2} + 2 b^{2} + 3 c^{2} + 4 ab + 12 b c + 6 c a$
D $a^{2} + 4 b^{2} + 9 c^{2}$

Step-by-Step Solution

Squares:

a^{2}, (2 b)^{2} = 4 b^{2}, (3 c)^{2} = 9 c^{2}

. Cross terms, each doubled:

2 (a) (2 b) = 4 ab, 2 (2 b) (3 c) = 12 b c, 2 (3 c) (a) = 6 c a

. So

(a + 2 b + 3 c)^{2} = a^{2} + 4 b^{2} + 9 c^{2} + 4 ab + 12 b c + 6 c a

Answer: A)

a^{2} + 4 b^{2} + 9 c^{2} + 4 ab + 12 b c + 6 c a

Key idea: Factorising polynomials using the factor theorem

To factorise a cubic like

p (x) = x^{3} - 6 x^{2} + 11 x - 6

:

**Step 1**: Try small integer values:

p (1) = 1 - 6 + 11 - 6 = 0

. ✓ So

(x - 1)

is a factor.
**Step 2**: Divide

p (x)

(x - 1)

using long division. Quotient:

x^{2} - 5 x + 6

.
**Step 3**: Factor the quadratic:

x^{2} - 5 x + 6 = (x - 2) (x - 3)

.
**Step 4**: Combine:

p (x) = (x - 1) (x - 2) (x - 3)

.

The zeros are 1, 2, 3 — three real zeros, as expected for a cubic.

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Using (x+y+z)2=x2+y2+z2+2xy+2yz+2zx, expand (a+2b+3c)2.

Step-by-Step Solution

Key idea: Factorising polynomials using the factor theorem

Master this chapter in 30 minutes a day.

More Class 9 Polynomials practice

Using $(x + y + z)^{2} = x^{2} + y^{2} + z^{2} + 2 x y + 2 yz + 2 z x$ , expand $(a + 2 b + 3 c)^{2}$ .